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10 tháng 7 2017

Theo đề bài ta có :

\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)

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

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

=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)

=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)

=> \(3x^3+5x-5x^2-x^4-2=0\)

=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)

=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)

=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)

=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)

=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)

=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)

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

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

Ta Thấy :

\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)

=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)

=> x = 1

ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)

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

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

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

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

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

Suy ra: \(x^2+5x+4=18\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-7;2}

22 tháng 3 2021

thank

9 tháng 3 2023

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\text{ĐKXĐ:}x\ne3;-1;\)

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

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

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

\(\Leftrightarrow2x^2-2x=4x\)

\(\Leftrightarrow2x^2-2x-4x=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\left(\text{loại}\right)\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0\right\}\)

9 tháng 3 2023

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne1;x\ne3\right)\\ \Leftrightarrow\dfrac{x.\left(x+1\right)+x.\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\\ \Rightarrow x^2+x+x^2-3x=4x\\ \Leftrightarrow2x^2-2x-4x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

loại

Vậy phương trình có tập nghiệm S={\(\varnothing\)}

28 tháng 1 2022

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

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

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

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

\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

 

 

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

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

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

\(\Leftrightarrow x=0\)

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

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

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

30 tháng 4 2021

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

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

\(\Rightarrow x-2-3\left(x-1\right)=-1\)

\(\Leftrightarrow x-2-3x+3=-1\)

\(\Leftrightarrow x-3x=-1+2-3\)

\(\Leftrightarrow-2x=-2\Leftrightarrow x=1\)

30 tháng 4 2021

đk: x\(\ne1,x\ne2\)

NV
18 tháng 3 2021

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

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

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

NV
18 tháng 3 2021

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

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

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

16 tháng 1 2022

 \(1,\dfrac{5x-1}{3}-1=2x+3\\ \Leftrightarrow\dfrac{5x-4}{3}=2x+3\\ \Leftrightarrow5x-4=3\left(2x+3\right)\\ \Leftrightarrow5x-4=6x+9\\ \Leftrightarrow6x+9-5x+4=0\\ \Leftrightarrow x+13=0\\ \Leftrightarrow x=-13\)

\(2,16x^2-3=\left(4x-3\right)\left(5x+1\right)\\ \Leftrightarrow16x^2-3=20x^2-15x+4x-3\\ \Leftrightarrow16x^2-3=20x^2-11x-3\\ \Leftrightarrow20x^2-11x-3-16x^2+3=0\\ \Leftrightarrow4x^2-11x=0\\ \Leftrightarrow x\left(4x-11\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{4}\end{matrix}\right.\)

\(3,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{-x\left(15-x\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}=\dfrac{x^2-15x}{x^2-4}\\ \Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=x^2-15x\)

\(\Leftrightarrow x^2-4x+4-3x-6-x^2+15x=0\\ \Leftrightarrow8x-2=0\\ \Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

9 tháng 4 2022

+ Pt thứ nhất :

Ta có mẫu thức chung là : \(2\left(x-3\right)\left(x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x\ne2\\x-3\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne3\\x\ne-1\end{matrix}\right.\)

Vậy \(ĐKXĐ\) là :\(x\ne2;3;-1\)

+ Pt thứ hai : 

Ta có mẫu thức chung là : \(\left(x-2\right)\left(x+3\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)

Vậy \(DKXD:\) \(\) \(x\ne2;-3\)

 

 

a: =>1+3x-6=-x+3

=>3x-5=-x+3

=>4x=8

=>x=2(loại)

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

=>3x-9+2x-4=x-1

=>5x-13=x-1

=>4x=12

=>x=3(loại)

c: =>x^2-2x+4+x^3+8=12

=>x^3+x^2-2x=0

=>x(x^2+x-2)=0

=>x(x+2)(x-1)=0

=>x=0 hoặc x=1

2 tháng 2 2023

tks yeu