\(2x^2\left(x-2\right)-2x\left(x-1\right)\left(x+1\right)=2x^3-4x^2-2x^3+2x=-4x^2+2x=-2x\left(2x-1\right)\)

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

\(=2x^3-4x^2-2x\left(x^2-1\right)\)

\(=2x^3-4x^2-2x^3+2x=-4x^2+2x\)

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

\(A=\left(2x-1\right)^2-5x\left(x-1\right)+2\left(x+1\right)\left(x-2\right)\)

\(=4x^2-4x+1-5x^2+5x+2\left(x^2-x-2\right)\)

\(=x^2-x-3\)

\((2x-1)(x-2)=2x^2-4x-x+2=2x^2-5x+2\)

2ax^2-a(1+2x^2)-[a-x(x+a)]

=2ax²-2ax²+a+x²-ax+a

=(2ax²-2ax²)-(a+a)+ax+x²

=0-2a+ax+x²

=x²+ax-2a

\(:P=5x(x-3)(x+3)-(2x-3)^2+34x(x+2)-5(x+2)^3+25x-1\)

\(P=5x(x^2-9)-(4x^2-12x+9)+34x^2+68x-5(x^3+6x^2+12x+8)+25-1\)

\(P=5x^3-45x-4x^2+12x-9+34x^2+68x-5x^3-30x^2-60x-40+25-1\)

\(P=(5x^3-5x^3)+(34x^2-4x^2-30x^2)+(12x-45x++68x+25x-60x)-(9+1)\)

\(P=-10\)

a:

ĐKXĐ: x<>2

|2x-3|=1

=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)

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

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

\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)

\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)

c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)

\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)

help voi a

a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)

\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)

\(=-19x^2+16x-4\)