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=>4x-6(2x+1)=2x-3x

=>4x-12x-6+x=0

=>-7x=6

hay x=-6/7

25 tháng 1 2022

\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)

\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)

\(\Leftrightarrow4x-12x-6=-x\)

\(\Leftrightarrow4x-12x-6+x=0\)

\(\Leftrightarrow-7x-6=0\)

\(\Leftrightarrow x=-\dfrac{6}{7}\)

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

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

1: Sửa đề: 2/x+2

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

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

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)

9 tháng 8 2023

câu b sai đề rồi anh ơi và câu a đâu rồi ạ

Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8+20x-12=0\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

4 tháng 3 2018

a) ĐKXĐ: \(x\ne\pm2\)

Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)

\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow x\left(x-2\right)=x^2+4\)

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

\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)

Vậy phương trình vô nghiệm

4 tháng 3 2018

b) ĐKXĐ: \(x\ne3;x\ne-1\)

Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=0\)

25 tháng 2 2021

`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`

`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`

`<=>-x-1-x+3=x^2+x-x^2+2x-1`

`<=>-2x+2=3x-1`

`<=>5x=3`

`<=>x=3/5`

Vậy `S={3/5}`

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

`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`

`<=>x+3-6x+12+6=0`

`<=>-5x+21=0`

`<=>x=21/5`

Vậy `S={21/5}`

a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

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

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

Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)

\(\Leftrightarrow3x-1=-2x+2\)

\(\Leftrightarrow3x+2x=2+1\)

\(\Leftrightarrow5x=3\)

hay \(x=\dfrac{3}{5}\)(nhận)

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

NV
21 tháng 4 2021

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)

Phương trình trở thành:

\(a^2-12b^2+ab=0\)

\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)

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

\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)

Bạn tự quy đồng và hoàn thành phần còn lại nhé

22 tháng 4 2021

e cảm ơn ạ

 

6 tháng 2 2018

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 ............

16 tháng 2 2018

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

16 tháng 2 2018

câu b lm tương tự nha