Bài 2:

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

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

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

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

Bạn cần viết lại đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Viết như thế này nhìn khó đọc quá.

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

a: ĐKXĐ: \(x\ne-2\)

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

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

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

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

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

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

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

Ta có: \(\dfrac{2x+3}{1-x^2}+\dfrac{2x+1}{x^2-2x+1}\)

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

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

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

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

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

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

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

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

Bài 4:

ĐKXĐ: \(x\ne0;x\ne\pm1\)

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

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

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

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

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

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

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

Kết bạn với tôi đi ,tôi cô đơn quá 

a: \(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)

\(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)

\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)

b: \(\dfrac{x+1}{2x-2}+\dfrac{x-1}{2x+2}+\dfrac{x^2}{1-x^2}\)

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

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

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

c: \(\dfrac{1}{x^2+xy}+\dfrac{2}{y^2-x^2}+\dfrac{1}{xy-x^2}\)

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

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

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

bằng 2^2 =4 

\(=\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}+\dfrac{2-3x}{\left(x+1\right)^2}\\ =\dfrac{\left(3x+2\right)\left(x+1\right)^2-6\left(x^2-1\right)+\left(2-3x\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\\ =\dfrac{10x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)