\(\left(x+2\right)\left(x-1\right)=x^2-x+2x-2=x^2+x-2\)

\(\frac{5x}{x-1}+\frac{5x}{x-1}=\frac{10x}{x-1}\)

\(\frac{1}{x-3}+\frac{-2}{x+3}+\frac{9-x}{x^2-9}=\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{9-x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x+3-2x+6+9-x}{\left(x-3\right)\left(x+3\right)}=\frac{-2x+18}{\left(x-3\right)\left(x+3\right)}\)

\(\frac{4x+8}{4-x^2}\left(x^2-2x\right)=\frac{4\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}x\left(x-2\right)\)

\(=-\frac{4x\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-4x\)

Thì ko được học dốt thế cũng hỏi :))

thì phải chăm ngoan học giỏi tập trung trong giờ học ko đc chwoi hoặc ăn trong giờ đặc biệt ko yêu đương 

\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\left(x-2\right)^2-\left(x^2-9\right)=6\)

\(x^2-4x+4-x^2+9=6\)

\(-4x+13=6\)

\(-4x=-7\)

\(x=\frac{7}{4}\)

\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow-4x+7=0\Leftrightarrow x=\frac{7}{4}\)

a + b + c = 0

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> a² + b² + c² = -2(ab + 2bc + 2ca)

=> (a² + b² + c²)² = [-2(ab + bc + ca)]²

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 4(a²b² + b²c² + c²a² + 2ab²c + 2a²bc + 2abc²) 

=> a⁴ + b⁴ + c⁴ = 4a²b² + 4b²c² + 4c²a² + 8a²bc + 8ab²c + 8abc² - 2a²b² - 2b²c² - 2a²c²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 8abc(a + b + c)

=> a⁴ + b⁴ + c⁴= 2a²b² + 2b²c² + c²a²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2abc(a + b + c) (Vì a + b + c = 0)

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2a²bc + 2ab²c + 2abc²

=>  a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + c²a² + a²bc + ab²c + abc²) 

=> a⁴ + b⁴ + c⁴ = 2(ab + bc + ca)² (đpcm)

\(2x^2-6x=2x\left(x-3\right)\)

\(2x^2-18=2\left(x^2-9\right)=2\left(x-3\right)\left(x+3\right)\)

\(x^3+3x^2+x+3=x\left(x^2+1\right)+3\left(x^2+1\right)=\left(x+3\right)\left(x^2+1\right)\)

\(x^2-y^2+6y-9=x^2-\left(y-3\right)^2=\left(x-y+3\right)\left(x+y+3\right)\)