\(\frac{x+2}{x-2}-\frac{6}{x+2}=\frac{x^2}{x^2-4}\)

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19 tháng 2 2020

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x+2}{x-2}-\frac{6}{x+2}=\frac{x^2}{x^2-4}\)

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{6}{x+2}-\frac{x^2}{x^2-4}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)^2-6\left(x-2\right)-x^2}{x^2-4}=0\)

\(\Leftrightarrow x^2+4x+4-6x+12-x^2=0\)

\(\Leftrightarrow-2x+16=0\)

\(\Leftrightarrow x=8\)

Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)

27 tháng 2 2020

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

a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)

\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)

\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)

\(\Leftrightarrow12x-10-8+2x=0\)

\(\Leftrightarrow10x-18=0\)

\(\Leftrightarrow10x=18\)

hay \(x=\frac{9}{5}\)

Vậy: \(x=\frac{9}{5}\)

b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)

\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)

\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)

\(\Leftrightarrow42-9x-2x+14=0\)

\(\Leftrightarrow56-11x=0\)

\(\Leftrightarrow11x=56\)

hay \(x=\frac{56}{11}\)

Vậy: \(x=\frac{56}{11}\)

c) ĐKXĐ: x∉{3;-3}

Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)

\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow6-x+2x-6=-5x-15\)

\(\Leftrightarrow x+5x+15=0\)

\(\Leftrightarrow6x=-15\)

hay \(x=\frac{-5}{2}\)(tm)

Vậy: \(x=\frac{-5}{2}\)

d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)

e) ĐKXĐ: x∉{4;-4}

Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)

\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)

\(\Leftrightarrow8x+10-4x+16=0\)

\(\Leftrightarrow4x+26=0\)

\(\Leftrightarrow4x=-26\)

hay \(x=\frac{-13}{2}\)(tm)

Vậy: \(x=\frac{-13}{2}\)

ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)

\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)

\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)

\(15-20x+6x-12=0\)

\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn 

5 tháng 5 2018
Sai đề câu a phải không
19 tháng 5 2017

Bạn không biết làm câu nào

22 tháng 5 2017

câu cuối bn

a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)

\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)

\(8x^3-46x^2-60x-180=0\)

=> vô nghiệm 

16 tháng 4 2020

b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)

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

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

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

\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)

=> 3x-6=0

<=> x=2 (ktm)

Vậy pt vô nghiệm

8 tháng 1 2020

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 :

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

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

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

\(\Leftrightarrow15x=3\)

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