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

a) (x - 1) (x² + x + 1) - (x + 1) (x² - x + 1) + 2(x - 1) (x + 1) - 2(x + 2)²

= x³ - 1 -  x³ - 1 + 2(x² - 1) - 2(x² + 4x + 4)

= -2 + 2x² -  2 - 2x² - 8x - 8

= -12  

a: Ta có: \(x\left(2-3x\right)+\left(3x^3-x^2\right):x\)

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

=x

b: Ta có: \(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)

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

\(=-2x^2\)

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: \(x\notin\left\{-1;2\right\}\)

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

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

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

a) \(\dfrac{2x}{x^2-6x+9}+\dfrac{x-2}{x-3}\) (ĐK: \(x\ne3\))

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

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

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

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

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

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

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

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

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

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

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

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

(1-x)^2-x(x-1)

Đặt \(x^2+1=a\)

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

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

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

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

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

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

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

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

\(=\dfrac{2x^4-2x^3+6x^2-3x+4}{\left(x^2+1\right)\left(x^2-x+1\right)}\)