a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)

\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)

\(=4x\cdot2y=8xy\)

b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)

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

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

Câu A) là \(\left(2x+y\right)^2-\left(y-2x\right)^2\)

Chứ ko phải là\(\left(2x+y\right)^2-\left(2x-y\right)^2\)

Nhưng dù sao thì cũng cảm ơn

\(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)

\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)

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

\(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)

\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)

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

\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)

\(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

a: ĐKXĐ: x<>1; x<>-1

TH1: \(x^2-1>0\)

=>\(x^2>1\)

=>x>1 hoặc x<-1

\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)

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

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

TH2: \(x^2-1<0\)

=>-1<x<1

\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)

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

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

b: \(B=2x:\frac12x+\left(x+1\right)^2\)

\(=\left(2:\frac12\right)\cdot\left(\frac{x}{x}\right)+x^2+2x+1\)

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

c: \(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)

\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)

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

d: \(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)

\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)

e: \(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)

\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)

f: \(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

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

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

(x+2)(x-2)-(x-3)(x+1)

=x^2-2x+2x-4-x^2-x-3x-3

=-4x-7

a)

=x²

=

=

b)

=

=

=

=

\(=x^3-27-4x^2+4x-1=x^3-4x^2+4x-28\)

thank you!

Đặt \(3x-1=y,x+2=z\)

\(\Rightarrow y^2-2yz+z^2=\left(y-z\right)^2\)

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

\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2=\left(\left(3x+1\right)-\left(3x-5\right)\right)^2=6^2=36\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)=\left(x^2-2y\right).5x^2\)

a. BT= ((3x+1) - (3x-5))²=6²=36

b. BT = (3x²-y-2x²-y). (3x²- y + 2x²+ y) = (x²-2y).5x²