Welcome to the world of Pokémon and matematics!
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x-1+(-1)x = x + 1
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\frac{2x+2(-2)+2}{2} + x = \frac{2}{-1-1}
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\frac{(-3x)3-3^2}{3}-3=3x+3
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\frac{4x(-4)^4}{4 \cdot 4 \cdot 4 \cdot 4} = 1
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\frac{5 \cdot 5 \cdot 5}{x7} = 5^2
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\frac{x}{x+x+x+x+x+x}+\frac{1}{6x} = \frac{1}{-6}
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\frac{(-7x)+7}{7+7}=\frac{1}{7}
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\frac{1}{8}\left( -\frac{1}{x} + \frac{1}{8} \right) = 0 \cdot 8
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9 \cdot 9^9(x-9) = \frac{0}{9}
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\frac{0+0x}{x + 0} = -(x^0 + x-0)
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\frac{1}{2}+ \frac{1}{3} + \frac{1}{12} + 1 = x
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x+(3x)(0 + 0)(x3)+x = 0^4-\frac{x}{2}
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\frac{x}{1+x}=3
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\frac{x^2}{(x+1)^2}=1
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\frac{x+2}{x}+\frac{1}{x} = -(-2)
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1^0 + 2^1-2^0 + 3^3-3^2 = \frac{1}{2}+\frac{x}{2}
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\frac{7}{x} = 5 -\left(4\frac{1}{2}\right)
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\frac{89 \cdot 2}{7} \cdot \frac{7 \cdot 1}{3 \cdot 89} = (1-2)x
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(x+1)(2+x)(3+1) - 4x^2 = 0
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\frac{2}{\left(\frac{1}{3x}\right)} = 1^{99}
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\frac{3}{\frac{1}{3x}-1}=(-1)^{99}
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\frac{\frac{(x+7)}{5} \cdot \frac{6}{(7+x)}}{25} = x
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\frac{-(x-2014)}{2014-x}\cdot x = -7
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\frac{2x+2}{1x+1} = x - 1 + \frac{2x}{x}
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x^2 = 25
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x^2 = 5x
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x^2 = 49
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x^2+2x+1 = 0
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(x-1)(x+0) = 0
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(x-1)(x+1) = 0
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(x+1)^2 = 0
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(5+x)(-7+6) = 0
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(x^2 -1)x -x^3 = 0
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x^2-99\cdot 99 = 0
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\frac{1}{x} = x
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3\cdot 3 \cdot 3 - 3^3 +4x = \frac{1}{x}
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\frac{2x+2}{x+1} -2 = x
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\frac{2(x+1)}{x+1}+\frac{x^2}{2} = 4
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\frac{x}{2}+\frac{1}{3}+\frac{2}{5} = 0
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\frac{1}{x^2} = 88 \cdot 88
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\frac{1}{2x+\frac{3}{4}} = 1
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6x-1^6x +1^0 = x25
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314 \cdot x^2 + 314 -17 \cdot 314 =0
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\frac{x+1}{2} \cdot \frac{2}{2^2} = 1
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\frac{1337}{1337^2x} = 1
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5x+0-9-(-2-1)(-1)=0
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\frac{(x+1)(x+1)}{(1-(-1))(-1)}=0
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\frac{1}{4x-5} = 1
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\frac{1}{x^2+2x+2} = 1
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5^{-1} + \frac{1}{5} = x
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\frac{1}{2^3}+\frac{1}{2\cdot2\cdot2}+2^{-3} = x
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5 \cdot 5^{-1} - x^2 = 0
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\frac{10}{10\cdot10\cdot10\cdot10} + \frac{10}{10^4} + 10\cdot10^{-4} - x = 0
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3^{4} \cdot 5^2 \cdot 3^{-3}\cdot 5^{-1}x =1
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\frac{2\cdot 7 \cdot 7}{7^{3}} - 2 \cdot 7^{2} \cdot 7^{-3} = -2^2 \cdot 7^{-1} +7x^2
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1^4 \cdot 1^{-9} \cdot 9^{7} \cdot 9^{-6}=-7 +x^2
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x^2 + x^2 = 2\cdot\left(\frac{7^4 \cdot 7^{-2}}{6^{-1}\cdot 6^3}\right)
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x(x-5) = 0
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\left(\frac{1}{\left(\frac{1}{5}\right)}\right)x -(5-x) + \left(\frac{1}{5}\right)^{-1} = 3
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\frac{3 \cdot 5^2 \cdot 1}{3^2 \cdot 5^2 \cdot(-1)}\cdot\frac{7}{7 \cdot 7}\cdot2^{-1} = x
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\frac{3 \cdot 7 \cdot 3 \cdot 14}{7^2 \cdot x \cdot 3^2} = \frac{1}{3}
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7^4 - 7 \cdot 7\cdot7\cdot7 + 3 \cdot 5 \cdot 3 \cdot 5 - (3 \cdot 5)^2 = x
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(3x)^2 + (3x)\cdot (3x) + 3\cdot x \cdot 3 \cdot x = 27
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\frac{3^4}{3\cdot3\cdot3\cdot3}+\frac{(7\cdot 8)^3}{7^3 \cdot 8^3} = x
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(3 \cdot x)^{-1} = 1
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-x^2-2x-1=0
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x7^{(5-3)}=14
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x\cdot x - x^2 + x^{-1} = 2014
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\frac{x}{7^3} = 7^{-3}
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\frac{6^{-15}}{x} = \frac{1}{6^{15}}
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\frac{7}{3x} = 21
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(x+1)(x+1^{-1})=0
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x(-x+81611)=0
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\frac{1234 \cdot 1234^{81249}}{1234^{81249}} = x
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\frac{1234^{81249+1}}{1234^{81249}} = x
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\frac{1234^{81249}}{1234^{81249-1}} = x
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x^2 + x -6 = 0
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(x-2)(x+3) = 0
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x^2-9x+20 = 0
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(x-5)(x-4) = 0
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x^2+7x+12 = 0
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(x+4)(x+3)=0
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12+13+14+15+16+17+18+19+20+21 = x
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-x-(-x-(-x))=3
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x^2 = 81
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(x-1)(3+2)(x-1) = 0
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x+2x+3x-\frac{1}{2}\left(\frac{14}{3}\right) = 0
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y-y+3x-2x = 5^{-1}
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2\cdot y\cdot(1-1^0)+x^{(2-1)} = \frac{5^{-1}}{1}
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(5^{-1} + 3^{-1})(3+5)x = 0
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\frac{(x-1)(x+9) -x^2-8x+9}{1+(32+44-324)^2 \cdot 91^{-2}} = x
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\frac{666-555}{111}(x-9^{-2})=0
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\left(6+\frac{x}{2}\right)\cdot \frac{2}{3(-1-(-1)+1)} = 0
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\frac{x+7}{4x+28}\cdot 6^{-1} \cdot 6 = x^2
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\frac{7}{2}(\frac{3x}{21}+\frac{4x}{28}) = 0
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5\cdot\frac{1}{2}+ \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1+1+1+1+1}{2}=x^{-1}
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(x+5)^2 = 36
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x^2 +2x - 1-1-1=0
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(x+3)(x-1)=0
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x^2-2x-35=0
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(x-7)(x+5)=0
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\frac{(x-11)(x+11)}{1} = -21
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(x+1)^2(3+1)=1
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\frac{7^1}{7^{-1}} = x
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\frac{(-7+x)^2}{11} = 0
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\frac{1}{\frac{1}{(x+1)}} + \frac{1}{\frac{1}{x+5}} = \frac{2}{3}
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\frac{1}{\frac{1}{x}+\frac{1}{x}}\cdot(2\cdot4)^{-1}=1
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\frac{1-7}{x+1} + \frac{2x+5}{x+1} = 1
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\frac{x+1}{(x+1)^2}+(x+1)^{-1} = 2
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\left(\frac{(x-3)\cdot(x-3)}{(x-3)^4} - (x-3)^{-2} \right)x = x+x
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x^2-\frac{5}{6}x + \frac{1}{6} = 0
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\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right) = 0
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\frac{-4}{x} + \frac{1}{x+1} = 1
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\frac{1}{x}+\frac{1}{2x} = 1
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-\frac{1}{x+2}-\frac{x}{1}=0
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(2-7)x+15-13+x=-3x+(-1\cdot(2-1))
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10x^2+5x = 0
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\frac{x^3}{x} = 16
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\frac{(3x+2x)(1-x)}{x} = 9
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(x-17357)(x+17357)=0
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3x\cdot(4+2x)=0
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3x\cdot\left(\frac{2}{3}+\frac{x}{2}\right) = 0
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(x+2)^2 -49 = 0
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-(-(-\left(2x+\frac{3}{2}\right)))=0
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x^{-2} = 3^{-2}
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\frac{1}{x^2} = 5^2
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5x-(3-2)x-(-1)^{2}=-x(-4)+x
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\frac{5 \cdot 5 \cdot 5 \cdot 3^7}{5^3 \cdot 3^6 \cdot x} = 1
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\frac{15}{(x-1)(x+3)}\cdot 0 +x^2 = \frac{9}{-1}(-1)
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\frac{4(4+4)^4-4}{(4+4)^4} + \frac{4}{(4+4)^4} = -x
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x = \frac{130}{1300}
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6x^2-13x+6 = 0
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4x+(-6+x)(-1)+3x=5
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\frac{2\cdot 3}{62x} = 1
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\frac{1}{2\cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = x
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\frac{2+3}{2\cdot 3} + \frac{3^2}{3+3} = x
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\frac{1}{-9^{-1}} + (x)^2 = 0
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x\cdot x = 3\cdot3 + 4\cdot 4
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x^2 = 5^2-4^2
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x^2=10\cdot\frac{1}{10^{-1}}
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\frac{5}{6} \cdot \frac{1}{\frac{15}{12}}x=-1
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\frac{\frac{4}{2+x}}{5} = 3
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z^{99}(x+(-1)^{99})=z^{99}
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(1+4)(\frac{1}{15}+x)=0
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(x-1)(2x+5)=0
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(x-5)^2 -3-5=0+1
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\frac{(x+1)^2}{(x+1)^3} = \frac{18}{2}
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\frac{4 \cdot 3 \cdot x}{3^3} = \frac{1}{x}
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5x^2-x+\frac{10x}{3}-\frac{3^{-1}}{2^{-1}} = 0
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\frac{7x+1}{-(x+5)(5+2)(-1)-34}+\frac{1+\frac{4}{x}}{\frac{3}{\frac{1}{2}}+1} = 0
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\frac{1}{x}+\frac{1}{x-1}=\frac{3x}{x(x-1)}
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