a ) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\) (1)

ĐKXĐ : x\(\ne1;-2.\)

\(\left(1\right)\Leftrightarrow x+2-7x+7=3\)

\(\Leftrightarrow-6x=-6\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy pt vô nghiệm .

b ) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)

Đặt \(x^2+2x+1=t\) ta được :

\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow6t^2+12t+6t^2+12t+6=7\left(t^2+3t+2\right)\)

\(\Leftrightarrow5t^2+3t-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{8}{5}\end{matrix}\right.\)

Khi t = 1

\(\Leftrightarrow\left(x+1\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Khi \(t=-\dfrac{8}{5}\)

\(\Leftrightarrow\left(x+1\right)^2=-\dfrac{8}{5}\) ( vô lí )

Vậy ............

mình lười nên nói cách làm nhé

B1: chuyển \(\dfrac{6}{x^2-9}\)sang vế trái và thêm dấu trừ ở trc \(\dfrac{6}{x^2-9}\)và vế phải =0

B2: để ý thấy \(x^2-9\)=(x-3).(x+3) tức là hằng đẳng thức số 3 ý

B3: quy đồng mẫu , mẫu số chung là (x-3).(x+3).(2x+7)

B4: chia cả hai vế cho (x-3).(x+3).(2x+7)

lưu ý : bước này là dấu⇒ chứ ko phải dấu ⇔ nhé

B5: giải pt như bình thg thui

ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{7}{2}\right\}\)

Ta có: \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\)

\(\Leftrightarrow\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

Suy ra: \(13x+39+x^2-9=12x+42\)

\(\Leftrightarrow x^2+13x+30-12x-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-4}

Đặt \(u=x^2-2x+2\)

=> Pt tương đương :

\(\dfrac{1}{u}+\dfrac{2}{u+1}=\dfrac{6}{u+2}\)

\(\Leftrightarrow\dfrac{\left(u+1\right)\left(u+2\right)+2u\cdot\left(u+2\right)}{u\left(u+1\right)\left(u+2\right)}=\dfrac{6u\left(u+1\right)}{u\left(u+1\right)\left(u+2\right)}\)

\(\Leftrightarrow\left(u+1\right)\left(u+2\right)+2u\left(u+2\right)=6u\left(u+1\right)\)

\(\Leftrightarrow u^2+3u+2+2u^2+4u=6u^2+6u\)

\(\Leftrightarrow-3u^2+u+2=0\)

\(\Rightarrow\left[{}\begin{matrix}u=1\\u=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2=1\\x^2-2x+2=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow x=1\)

Kết luận \(x=1\)

\(pt\Leftrightarrow\dfrac{1}{\left(x-1\right)^2+1}+\dfrac{2}{\left(x-1\right)^2+2}=\dfrac{6}{\left(x-1\right)^2+3}\)

Đặt: \(\left(x-1\right)^2=t\ge0\)

\(pt\Leftrightarrow\dfrac{1}{t+1}+\dfrac{2}{t+2}=\dfrac{6}{t+3}\)

\(\Rightarrow\dfrac{t+2+2\left(t+1\right)}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)

\(\Rightarrow\dfrac{t+2+2t+2}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)

\(\Rightarrow\dfrac{3t+4}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)

\(\Rightarrow\left(3t+4\right)\left(t+3\right)=6\left(t+1\right)\left(t+2\right)\)

Phân tích ra:v

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

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

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

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

=>x=0(nhận) hoặc x=3(loại)

đk : x khác -1 ; 3 

\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x\Leftrightarrow2x^2-2x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\left(ktm\right)\)

b) Đặt \(x^2+2x+3=a\)(a>0)

Ta có: \(\dfrac{x^2+2x+7}{\left(x+1\right)^2+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+1+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+3}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{a+4}{a}=a+1\)

\(\Leftrightarrow a^2+a=a+4\)

\(\Leftrightarrow a^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-2\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

ĐKXĐ của cả 2 pt trên đều là `x in RR`

`a,1/(x^2-2x+2)+2/(x^2-2x+3)=6/(x^2-2x+4)`

Đặt `a=x^+2x+3(a>=2)` ta có:

`1/(a-1)+2/a=6/(a+1)`

`<=>a(a+1)+2(a-1)(a+1)=6a(a-1)`

`<=>a^2+a+2(a^2-1)=6a^2-6a`

`<=>a^2+a+2a^2-2=6a^2-6a`

`<=>3a^2-5a+2=0`

`<=>3a^2-3a-2a+2=0`

`<=>3a(a-1)-2(a-1)=0`

`<=>(a-1)(3a-2)=0`

`a>=2=>a-1>=1>0`

`a>=2=>3a-2>=4>0`

Vậy pt vô nghiệm

`(x^2+2x+7)/((x+1)^2+2)=x^2+2x+4`

`<=>(x^2+2x+7)=(x^2+2x+4)(x^2+2x+3)`

Đặt `a=x^2+2x+3(a>=2)`

`pt<=>a+4=a(a+1)`

`<=>a^2+a=a+4`

`<=>a^2=4`

`<=>a=2` do `a>=2`

`<=>(x+1)^2+2=2`

`<=>(x+1)^2=0`

`<=>x=-1`

Vậy `S={-1}`

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)