a. 5x + 3(x² - x - 1)

= 5x + 3x² - 3x - 3

= 3x² + 5x - 3x - 3

= 3x² + 2x - 3

b. (5 - x)(5 + x) - (2x - 1)²

25 - x² - (4x² - 4x + 1)

= 25 - x² - 4x² + 4x - 1

= 25 - 1 - x² - 4x² + 4x 

= 24 - 5x² + 4x

a) \(5x+3\left(x^2-x-1\right)=5x+3x^2-3x-3=3x^2+2x-3\)

b) \(\left(5-x\right)\left(5+x\right)-\left(2x-1\right)^2=25-x^2-4x^2+4x-1=-5x^2+4x+24\)

a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)

b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)

c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)

b,

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

c,

\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)

\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

a) \(=8-x^3-x\left(16-x^2\right)=8-x^3-16x+x^3=-16x+8\)

b) \(=\left[\left(x+3\right)\left(x^2-3x+9\right)\right]:\left(x^2-3x+9\right)-x+7\)

\(=x+3-x+7=10\)

Dễ

 Thế

Mà

Cũnhoir

Dc

Ạ

Chịu

Chắc

Phải

Ngu 

Lamqs

Mới

Hỏi

Câu

Này

D

TỪ BẠN ƠI

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$

$=-3x^2y+3xy^2=3xy(y-x)$

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)

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

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

b: \(\frac{x+1}{x^2-2x-8}\cdot\frac{4-x}{x^2+x}\)

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

c: \(\frac{9x+5}{2\left(x-1\right)\left(x+3\right)^2}-\frac{5x-7}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{9x+5-5x+7}{2\left(x-1\right)\left(x+3\right)^2}=\frac{4x+12}{2\left(x-1\right)\left(x+3\right)^2}\)

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

d: \(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)

\(=\frac{18}{\left(x+3\right)\left(x-3\right)^2}-\frac{3}{\left(x-3\right)^2}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{18-3x-9-x^2+3x}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{-x^2+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-1}{x-3}\)

e: \(\frac{1}{x^2-x+1}+\frac{1}{1-x^2}+\frac{2}{x^3+1}\)

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

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

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

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^4-10x^3+6x^2\)

c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)

d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)