a) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{1\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

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

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

\(\Leftrightarrow\frac{3x}{x^2+x+1}=0\)

=> 3x=0

<=> x=0 (tmđk)

đọc cách lm trrong sbt nha bạn lớp 8

mk học lớp 6 lên 7

nâng cao

các bạn giúp mình với. cảm ơn 

giúp mình với

a,\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)\(\Leftrightarrow\frac{13\left(x+3\right)}{\left(x^2-9\right)\left(2x+7\right)}+\frac{x^2-9}{\left(x^2-9\right)\left(2x+7\right)}-\frac{6\left(2x+7\right)}{\left(x^2-9\right)\left(2x+7\right)}=0\)

\(\Leftrightarrow x+x^2-12=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=3\end{cases}}\)

b,\(\frac{x-3}{x-5}+\frac{1}{x}=\frac{x+5}{x\left(x-5\right)}\Leftrightarrow\frac{x\left(x-3\right)}{x\left(x-5\right)}+\frac{x-5}{x\left(x-5\right)}-\frac{x+5}{x\left(x-5\right)}=0\)

\(\Leftrightarrow x^2-3x-10=0\Rightarrow\orbr{\begin{cases}x=5\left(L\right)\\x=-2\end{cases}}\)

c,\(\frac{1}{x+2}+\frac{1}{x\left(x-2\right)}-\frac{8}{x\left(x^2-4\right)}=0\)\(\Leftrightarrow\frac{x\left(x-2\right)}{x\left(x^2-4\right)}+\frac{x+2}{x\left(x^2-4\right)}-\frac{8}{x\left(x^2-4\right)}=0\)

\(\Leftrightarrow x^2-x-6=0\Rightarrow\orbr{\begin{cases}x=3\\x=-2\left(L\right)\end{cases}}\)

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

\(\Leftrightarrow-x^2-5x-10=0\)(vô nghiệm)

\(\)

13(x+3)+(x+3)(x-3)=6(2x+7)

13x+39+x^2-9-12x-42=0

x^2+x-12=0

x=3 và x=-4

**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Theo bài ra , ta có :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)

Quy đồng và khử mẫu phương trình ta đk :

\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)

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

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

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

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

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

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

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

Kết hợp với ĐKXĐ ta có : x = -4

Vậy \(S=\left\{-4\right\}\)

Chúc bạn học tốt =))

ĐKXĐ: x\(\ne\)3;-7/2;-3

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)

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

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

Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4

Đk: x khác 3; -3 và -7/2

Từ đề suy ra:

13/(x-3)(2x+7) + 1/2x+7 - 6/(x-3)(x+3) = 0

<=> \(\frac{13\left(x+3\right)+\left(x-3\right)\left(x+3\right)-6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=0\)

<=> 13(x+3) + (x-3)(x+3) + 6(2x+7) = 0

<=> 13x + 39 + x² - 9 + 12x + 42 = 0

<=> x² + 25x + 72 = 0

<=> (x + 25/2)² = 337/4

<=> \(\left[\begin{matrix}x+\frac{25}{2}=\sqrt{\frac{337}{4}}\\x+\frac{25}{2}=-\sqrt{\frac{337}{4}}\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=\frac{-25+\sqrt{337}}{2}\\x=\frac{25-\sqrt{337}}{2}\end{matrix}\right.\)(TM)

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