Bài 1: (Sgk/36):

a. \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\) vì

5y . 28x = 140xy

7 . 20xy = 140xy

=> 5y . 28x = 7 . 20xy

Vậy \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\)

b. \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\) vì

3x . 2(x+5) = 6x²+30x

2 . 3x(x+5) = 6x²+30x

=> 3x . 2(x+5) = 2 . 3x(x+5)

Vậy \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\)

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

(x+2) (x²-1) = (x+2) (x-1) (x-1)

=> (x+2) (x²-1) = (x-1) (x+2) (x+1)

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

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

(x-1) (x²-x-2) = x³-2x²-x+2

(x+1) (x²-3x+2) = x³-2x²-x+2

=> (x-1) (x²-x-2) = (x²-3x+2) (x+1)

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

a. \(x^2y^3.35xy=5.7x^3y^4\)

\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)

\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)

\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)

\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)

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

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

\(\Rightarrowđpcm\)

\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)

\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)

Xin được mạn phép chữa đề.

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

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

Vậy.......................

1: Sai

2: Sai

3: Đúng

4: Đúng

Lời giải:

1. Đúng

2. Sai

3. Đúng

4. Đúng

\(a,VP=\dfrac{x^2+4x+3}{x^2+6x+9}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x+1}{x+3}=VT\)

Vậy ta có đpcm 

b, \(VP=\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}=\dfrac{x+y}{3x}=VT\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{x^2+4x+3}{x^2+6x+9}\)

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

\(=\dfrac{x+1}{x+3}\)

b: Ta có: \(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)

\(=\dfrac{3x\left(x+y\right)\left(x+y\right)}{3x\cdot3x\cdot\left(x+y\right)}\)

\(=\dfrac{x+y}{3x}\)

\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)

\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)

\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)

Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:

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

\(=\dfrac{y^2-1+1}{y}\)

\(=\dfrac{y^2}{y}\)

\(=y\)

\(=x^2+7x+11\)

Vậy \(N=x^2+7x+11\).

\(\text{#}Toru\)

a/ ĐK: $x\ne -5$

$\dfrac{6x^2+30x}{4}=\dfrac{6x(x+5)}{4}=\dfrac{3x(x+5)}{2}$ 

Đề này sai

b/ ĐK: $x\ne \pm 1$

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

$\to$ ĐPCM

Câu a sai đề nhé.

a, Xét \(VT=\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)

\(VP=\dfrac{6x^2+30x}{4}=\dfrac{6x\left(x+5\right)}{4}=\dfrac{3x\left(x+5\right)}{2}\)

Vậy \(VT\ne VP\)hay đpcm ko xảy ra 

b, \(VP=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}=VT\)

Vậy ta có đpcm 

\(a,ĐK:x\ne0;x-1\ne0\Leftrightarrow x\ne0;x\ne1\\ b,ĐK:x^2-4=\left(x-2\right)\left(x+2\right)\ne0\Leftrightarrow x\ne2;x\ne-2\)

a: ĐKXĐ: \(x\notin\left\{0;1\right\}\)

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

\(4x\left(x+2\right)^3\)

Mtc:

8x(x+2)^3