a, \(\left(2x-1\right)\left(3-2x\right)=6x-4x^2-3+2x=-4x^2+8x-3\)

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

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

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a: =9x^2-12x+4-4x^2+14x

=5x^2+2x+4

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

a) 3x.(x² - 2)

= 3x.x² + 3x.(-2)

= 3x³ - 6x

b) (6x³ + 2x² - 4x) : 2x

= 6x³ : 2x + 2x² : 2x - 4x : 2x

= 3x² + x - 2

c) 2x(x² - 1)

= 2x.x² - 2x.1

= 2x³ - 2x

a) \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)

b) \(\left(x+3\right)^2+\left(x-1\right)\left(3+2x\right)=x^2+6x+9+3x+2x^2-3-2x\)

\(=3x^2+7x+6\)

a: \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)

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

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

\(=3x^2+7x+6\)

bằng 2^2 =4 

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

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

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

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

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

thực ra mình cũng cố rồi nhưng mà IQ có hạn nên nghĩ mãi ko ra, thế nên mới phải cầu cứu mấy bạn giỏi hơn đấy =)

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: \(x\notin\left\{-1;2\right\}\)

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

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

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