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28 tháng 1 2020

\(ĐKXĐ:x\ne-1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow x-2-5x-5=15\)

\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)

Vậy \(S=\left\{\frac{-11}{2}\right\}\)

28 tháng 1 2020

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)

\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow-4x-7=15\)

\(\Leftrightarrow-4x=22\)

\(\Leftrightarrow x=22:\left(-4\right)\)

\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)

Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)

9 tháng 8 2015

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

<=>\(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+2}+\frac{1}{x^2+2}-\frac{1}{x^2+3}+...+\frac{1}{x^2+4}-\frac{1}{x^2+5}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)

<=>(x2+5)-(x2+1)=-(x2+1)(x2+5)

<=>4=-x4-6x2-5

<=>x4+6x2+9=0

<=>(x2+3)2=0

<=>x2+3=0

Do x2>0

=>x2+3>0 nên PT vô nghiệm

25 tháng 2 2019

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

25 tháng 2 2019

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

19 tháng 4 2020
https://i.imgur.com/wgXaoMx.jpg
21 tháng 7 2019

\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)

\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)

\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)

\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)

\(\Rightarrow-16x+12x+16-3=24x+48-18\)

\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)

21 tháng 7 2019

nhung sao lai binh phuong len vay

-------------------ko chép đề nha---------

\(\Leftrightarrow\frac{4\left(x^2-4x+4\right)-3\left(2x+1\right)}{12}=\frac{12-2\left(4x^2-12x+9\right)}{12}\)

\(\Rightarrow4x^2+16x+16-6x-3=12-8x^2+24x-18\)

\(\Leftrightarrow4x^2+10x+13=-8x^2+24x-6\)

\(\Leftrightarrow4x^2+8x^2+10x-24x+13+6=0\)

\(\Leftrightarrow12x-14x+19=0\)

Ta có :\(\Delta'=7^2-12.19=-179< 0\)

\(\Rightarrow\)phương trình vô nghiệm

21 tháng 7 2019

\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)

<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)

<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)

<=> \(3x^4+18x^2+12x-33=0\)

<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)

<=> \(x-1=0\)

<=> \(x=1\)

Mà vì: \(x^3+x^2+7x+11\ne0\)

=> x = 1

21 tháng 7 2019

\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)

\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)

\(=>\left(x^2+3\right)^2=16\)

Đến đây tự tìm nha 

 Hok tốt 

9 tháng 3 2019

\(\left(x+5\right)+\left(x-5\right)+5x+x\div5=180\)

\(\Leftrightarrow\left(x+x+5x\right)+\left(5-5\right)+\frac{x}{5}=180\)

\(\Leftrightarrow7x+0+\frac{x}{5}=180\)

\(\Leftrightarrow7x+\frac{x}{5}=180\)

\(\Leftrightarrow\frac{35x+x}{5}=180\)

\(\Leftrightarrow35x+x=180.5\)

\(\Leftrightarrow36x=900\)

\(\Leftrightarrow x=\frac{900}{36}\)

\(\Leftrightarrow x=25\)

Vậy phương trình có 1 nghiệm duy nhất là 25

9 tháng 3 2019

(x + 5) + (x - 5) + 5x + \(\frac{x}{5}\)= 180

<=> x + 5 + x - 5 + 5x + \(\frac{x}{5}\) = 180

<=> 7x + \(\frac{x}{5}\) = 180

<=> \(\frac{36x}{5}=180\)

\(\Leftrightarrow x=\frac{180.5}{36}=25\)