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

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

<=> x²+2x-x+2=2

<=> x²+x=2-2

<=> x²+x=0

<=>x(x+1)=0

<=>x=0 hoặc x+1=0

<=>x=0 hoặc x = -1

a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

<=> x-3 =10x-15

<=> x-10x= -15+3

<=> -9x = -12

<=> x = \(\frac{-12}{-9}\)

<=> x = \(\frac{4}{3}\)

\(\left(x-1\right)\left(x+1\right)-2\left(2x+3\right)\le\left(x-2\right)^2+x\)

\(\Leftrightarrow x^2-1-4x-6\le x^2-4x+4+x\)

\(\Leftrightarrow x^2-4x-7\le x^2-3x+4\)

\(\Leftrightarrow x^2-4x-x^2+3x\le7+4\)

\(\Leftrightarrow-x\le11\)

\(\Leftrightarrow x\le-11\)

biết đừng đăng anh à

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

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

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

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x²+2)

<=> x²+3x+2+x²-3x+2 = 2x²+4

<=> x²+x²-2x²+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)²-4(x²-2x+1) = 0

<=> x²+2x+1-4x²+8x-4 = 0

<=> -3x²+10x-3 = 0

giải phương trình

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

c) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=1\) đặt x-2 =t " cho bé hệ số lại

ĐK : \(\left\{\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t\ne0\\t\ne-2\end{matrix}\right.\)

\(\frac{t-1}{t}=\frac{t}{t-2}\Leftrightarrow\left(t-1\right)\left(t-2\right)=t^2\Leftrightarrow t^2-3t+2=t^2\Rightarrow-3t=-2\)

\(t=\frac{2}{3}\Rightarrow x=2+\frac{2}{3}=\frac{8}{3}\)

a) \(A=\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3x}\) xem lại đề thấy cái mẫu VP vô duyên thế!

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

\(B=\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(4x^2-1\right)}{MSC}=\frac{\left(2x^2+2x+2\right)+\left(2x^2+x-3\right)-4x^2+1}{MSC}=0\)

\(B=0\Leftrightarrow\frac{3x}{MSC}=0=>x=0\) thảo mãn đk x khác 1

Kết luận: x=0 là nghiệm duy nhất.

a)

ĐKXĐ: \(x\neq 0; x\neq -10\)

\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)

\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)

\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)

\(\Rightarrow 12(2x+10)=x(x+10)\)

\(\Leftrightarrow x^2-14x-120=0\)

\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)

b)

ĐKXĐ: \(x\neq 0; x\neq 3\)

PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)

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

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

\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$

c)

ĐKXĐ: \(x\neq \pm 2\)

\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)

\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)

\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)

\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)

Vậy pt vô nghiệm.

d)

ĐKXĐ: \(x\neq -2; x\neq 3\)

PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)

\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)

\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)

Vậy..........