Bài 3: Rút gọn phân thức

\(a,\dfrac{x^2+6x+9}{x+3}\\ đk:x\ne-3\\ =\dfrac{\left(x+3\right)^2}{x+3}=x+3\)

b, Thay \(x=-2\left(t/mđk\right)\) vào 

\(-2+3=1\)

Vậy tại \(x=-2\) thì biểu thức = 1 

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

\(A=\dfrac{x^2+2.x.3+3^2}{x+3}\)

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

\(A=x+3\)

b) Thay x = -2 vào A ta được A = -2 + 3 = 1

Vậy khi x = -2 thì A = 1

\(a)\dfrac{x^2+6x+9}{x+3}=\dfrac{\left(x+3\right)^2}{x+3}=x+3\)

\(\text{b)Thay x=-2 vào biểu thức x+3,ta được:}\)

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

\(\text{Vậy giá trị của biểu thức trên tại x=-2 là:1}\)

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

\(\dfrac{8x^2+16x^2+8x}{4x^2+4x}\)

= \(\dfrac{24x^2+8x}{4x^2+4x}\)

= \(\dfrac{4x(6x+2)}{4x(x+1)}\)

= \(\dfrac{6x+2}{x+1}\)

\(\dfrac{8x^2+16x^2+8x}{4x^2+4x}\\ =\dfrac{8x\left(x+2x+1\right)}{4x\left(x+1\right)}\\ =\dfrac{2\left(x+2x+1\right)}{x+1}\)

\(\left(2x-3\right)\left(4x^2+6x+9\right)-4x\left(2x^2-1\right)\)

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

giúp mik vs ạ

help

a. \(M=\dfrac{11x^2+11x}{x^2-1}\left(đk:x\ne\pm1\right)\)

\(=\dfrac{11x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{11x}{x-1}\)

b. \(\dfrac{11x}{x-1}=x.\dfrac{11}{x-1}\)

\(M\in Z\Leftrightarrow x.\dfrac{11}{x-1}\in Z\Leftrightarrow\dfrac{11}{x-1}\in Z\) (vì x nguyên)

\(\Leftrightarrow x-1\inƯ_{\left(11\right)}=\left\{\pm1;\pm11\right\}\)

+) \(x-1=1\Leftrightarrow x=2\)

+) \(x-1=-1\Leftrightarrow x=0\)

+) \(x-1=11\Leftrightarrow x=12\)

+) \(x-1=-11\Leftrightarrow x=-10\)

b,\(=\dfrac{3x}{4y^3}\)

c,=\(\dfrac{x-z}{2}\)

a,=\(\dfrac{-3x\left(x-1\right)}{2\left(x-1\right)}=\dfrac{-3x}{2}\)

Bài 3:

a: \(\Leftrightarrow2x^2-10x+x-5-2x^2=-10\)

=>-9x-5=-10

=>-9x=-5

=>x=5/9

b: \(\Leftrightarrow\left(x-5-7\right)\left(x-5+7\right)=0\)

=>(x-12)(x+2)=0

=>x=12 hoặc x=-2

Bài 2:

a: =5x*x-5x*3

=5x(x-3)

b: =(x+3)^2-4y^2

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

Bài 2:

a: \(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

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

b: Để C là số nguyên thì \(2x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{0;\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

Bài 3:

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

\(=\dfrac{-x^2+4+x^2-9+2-x}{\left(x+3\right)\left(x+2\right)}\cdot\dfrac{x-1}{-1}\)

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

b: Để A=0 thì x-1=0

=>x=1(loại)

c: Để A>0 thì x-1/x+2>0

=>x>1 hoặc x<-2

\(\dfrac{\left(x-2\right)^2\left(2x+2x^2\right)}{\left(x+1\right)\left(4x-x^3\right)}\\ =\dfrac{\left(x-2\right)^22x\left(1+x\right)}{\left(x+1\right)x\left(4-x^2\right)}\\ =\dfrac{2\left(x-2\right)^2}{\left(2-x\right)\left(2+x\right)}\\ =\dfrac{2\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\\ =\dfrac{2\left(2-x\right)}{2+x}\)