Bài 1:

a) ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow \frac{x(x-1)+3(x+3)}{3x}=2$

$\Rightarrow x^2+2x+9=6x$

$\Leftrightarrow x^2-4x+9=0$

$\Leftrightarrow (x-2)^2=-5<0$ (vô lý)

Do đó pt vô nghiệm

b) 

$x^2-25=(2x-1)(x+5)$

$\Leftrightarrow (x-5)(x+5)=(2x-1)(x+5)$

$\Leftrightarrow (x-5)(x+5)-(2x-1)(x+5)=0$

$\Leftrightarrow (x+5)[(x-5)-(2x-1)]=0$

$\Leftrightarrow (x+5)(-x-4)=0$

$\Rightarrow x+5=0$ hoặc $-x-4=0$

$\Rightarrow x=-5$ hoặc $x=-4$

c) ĐKXĐ: $x\neq 0; x\neq -2$

PT $\Leftrightarrow \frac{x(x-2)}{x(x+2)}-\frac{x^2+2}{x(x+2)}=\frac{3(x+2)}{x(x+2)}$

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

$\Leftrightarrow -8=5x$

$\Leftrightarrow x=\frac{-8}{5}$ (thỏa mãn)

Bài 2:

a) 
ĐKXĐ: \(\left\{\begin{matrix} x-3\neq 0\\ 9-x^2\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 3\)

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

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

b) 

\(M=\frac{x^2}{x-3}=\frac{x^2-9+9}{x-3}=x+3+\frac{9}{x-3}\)

Với $x$ nguyên, để $M$ nguyên thì $\frac{9}{x-3}$ nguyên

Với $x$ nguyên, để $\frac{9}{x-3}$ nguyên thì $x-3$ là ước của $9$

$\Rightarrow x-3\in\left\{\pm 1;\pm 3;\pm 9\right\}$

$\Rightarrow x\in\left\{2;4;6;0;12;-6\right\}$ (đều thỏa)

Bài 6: 

a) Ta có: \(4x-10=0\)

\(\Leftrightarrow4x=10\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)

b) Ta có: \(7-3x=9-x\)

\(\Leftrightarrow-3x+7-9+x=0\)

\(\Leftrightarrow-2x-2=0\)

\(\Leftrightarrow-2x=2\)

\(\Leftrightarrow x=-1\)

Vậy: S={-1}

c) Ta có: \(2x-\left(3-5x\right)=4\left(x+3\right)\)

\(\Leftrightarrow2x-3+5x=4x+12\)

\(\Leftrightarrow7x-3-4x-12=0\)

\(\Leftrightarrow3x-15=0\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

Vậy: S={5}

d) Ta có: \(5-\left(6-x\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow x+11-12+8x=0\)

\(\Leftrightarrow9x-1=0\)

\(\Leftrightarrow9x=1\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy: \(S=\left\{\dfrac{1}{9}\right\}\)

e) Ta có: \(4\left(x+3\right)=-7x+17\)

\(\Leftrightarrow4x+12+7x-17=0\)

\(\Leftrightarrow11x-5=0\)

\(\Leftrightarrow11x=5\)

\(\Leftrightarrow x=\dfrac{5}{11}\)

Vậy: \(S=\left\{\dfrac{5}{11}\right\}\)

bn cs the giups mk cacs bt cons laij dc ko tu bt 1- 5

1.

PT $\Leftrightarrow (x^2+2xy+y^2)-(y^2+6y+9)=0$

$\Leftrightarrow (x+y)^2-(y+3)^2=0$

$\Leftrightarrow (x+y-y-3)(x+y+y+3)=0$

$\Leftrightarrow (x-3)(x+2y+3)=0$

$\Rightarrow x-3=0$ hoặc $x+2y+3=0$

Nếu $x-3=0\Leftrightarrow x=3$. Vậy $(x,y)=(3,a)$ với $a$ nguyên bất kỳ.

Nếu $x+2y+3=0\Leftrightarrow x=-2y-3$ lẻ. Vậy $(x,y)=(-2a-3,a)$ với $a$ nguyên bất kỳ.

2. 

PT $\Leftrightarrow x^2=(y^2+2y+1)+12$

$\Leftrightarrow x^2=(y+1)^2+12\Leftrightarrow x^2-(y+1)^2=12$

$\Leftrightarrow (x-y-1)(x+y+1)=12$
Vì $x-y-1, x+y+1$ là số nguyên và cùng tính chẵn lẻ nên xảy ra các TH sau:

TH1: $x-y-1=2; x+y+1=6\Rightarrow x=4; y=1$

TH2: $x-y-1=6; x+y+1=2\Rightarrow x=4; y=-3$

TH3: $x-y-1=-2; x+y+1=-6\Rightarrow x=-4; y=-3$

TH4: $x-y-1=-6; x+y+1=-2\Rightarrow x=-4; y=1$

a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)

b: =>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

V...\(S=\left\{2010\right\}\)

^^

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

tích mình , rồi mình làm cho

a) 7x + 21 = 0 <=> 7x = -21 <=> x=-3

b) 5x - 2 = 0 <=> 5x =2 <=> x= 2/5

c) 12 - 6x =0 <=>6x = -12 <=> x= -2 

d) -2x + 14 = 0 <=> 2x = 14 <=> x = 7

a) 7x + 21 = 0

<=> 7x = -21

<=> x = -3

Vậy S = {-3}

b) 5x - 2 = 0

<=> 5x = 2

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy:....

c) 12 - 6x =0 

<=> 6x = 12

<=> x = 2

Vậy S = {2}

d) -2x + 14 = 0

<=> -2x = -14

<=> x = 7

Vậy S = {7}

Đk: \(\hept{\begin{cases}x^2-9\ge0\\2x-6+\sqrt{x^2-9}\ne0\end{cases}}\)

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

TH1: \(\hept{\begin{cases}x+3\ge0\\x-3\ge0\end{cases}\Leftrightarrow}x\ge3\)

\(A=\frac{\sqrt{x+3}.\sqrt{x+3}+2\sqrt{x-3}.\sqrt{x+3}}{2\sqrt{x-3}\sqrt{x-3}+\sqrt{x+3}.\sqrt{x-3}}\)

\(A=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}=\frac{\sqrt{x^2-9}}{x-3}\)

TH2: \(\hept{\begin{cases}x+3\le0\\x-3\le0\end{cases}\Leftrightarrow}x\le-3\)

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

\(A=\frac{\sqrt{-x-3}\left(\sqrt{-x-3}+2\sqrt{-x+3}\right)}{\sqrt{-x+3}\left(2\sqrt{-x+3}+\sqrt{-x-3}\right)}=\frac{\sqrt{-x-3}}{\sqrt{-x+3}}=\frac{\sqrt{x^2-9}}{3-x}\)