\(\frac{12}{x-1}-\frac{8}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\frac{12\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{8\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\) \(\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\left(12x+12\right)-\left(8x-8\right)=x^2-1\)

\(\Leftrightarrow12x+12-8x+8=x^2-1\)

\(\Leftrightarrow12x+12-8x+8-x^2+1=0\)

\(\Leftrightarrow-x^2+4x+21=0\)

\(\Leftrightarrow x^2-4x-21=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)-25=0\)

\(\Leftrightarrow\left(x-2\right)^2-5^2=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{7;-3\right\}\)

a thiếu

chỗ x phải có chữ thỏa mãn nữa nha

sorry sora cưng

\(\frac{x^3+7x^2+6x-30}{x^3-1}=\frac{x^2-x+16}{x^2+x+1}\) \(\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow\frac{x^3+7x^2+6x-30}{\left(x-1\right)\left(x^2+x+1\right)}=\) \(\frac{\left(x^2-x+16\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^3+7x^2+6x-30=x^3-2x^2+17x-16\)

\(\Leftrightarrow9x^2-11x-14=0\)

\(\Leftrightarrow\left(9x^2-11x+\frac{121}{36}\right)-\frac{625}{36}=0\)

\(\Leftrightarrow\left(3x-\frac{11}{6}\right)^2-\left(\frac{25}{6}\right)^2=0\)

\(\Leftrightarrow\left(3x-6\right)\left(3x+\frac{7}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=\frac{-7}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{7}{9}\left(tm\right)\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{2;\frac{-7}{9}\right\}\)

\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)

\(\Rightarrow\frac{x^2-8}{\left(x+4\right)\left(x-4\right)}=\frac{x-4}{\left(x+4\right)\left(x-4\right)}+\frac{x+4}{\left(x-4\right)\left(x+4\right)}\)

\(\Rightarrow x^2-8=x-4+x+4\)

\(\Rightarrow x^2-8=2x\)

\(\Rightarrow x^2-2x-8=0\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4.1.\left(-8\right)=4+32=36>0\)

phương trình có 2 nghiệm phân biệt : \(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{2+\sqrt{36}}{2}=\frac{2+6}{2}=\frac{8}{2}=4\)

                                                      \(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{2-\sqrt{36}}{2}=\frac{2-6}{2}=\frac{-4}{2}=\left(-2\right)\)

a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)

\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)

\(\Leftrightarrow96x+744=-6x+48\)

\(\Leftrightarrow102x=-696\)

\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)

Vậy .....

b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)

\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)

\(\Leftrightarrow x=-5\) (nhận)

Vậy ....

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)

\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)

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

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

\(\Leftrightarrow-x^2-3=3x-3\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\) 

Vậy x = 0 

\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)

\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow-6x+3=8\)

\(\Leftrightarrow x=-\frac{5}{6}\)

Vậy ... 

Ai muốn kết bn ko!

Tiện thể tk mình luôn nha!

Konasuba

\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\)
\(ĐKXĐ:x^2+2x-3=\left(x+1\right)\left(x-3\right)\\ \Rightarrow x\ne-1;x\ne3\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x^2\)
\(\Leftrightarrow x^2+x+x^2-3x=4x^2\)
\(\Leftrightarrow2x^2-2x=4x^2\)
\(\Leftrightarrow2x^2-4x^2-2x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow2x\left(-x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\-x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(N\right)\\x=-1\left(L\right)\end{cases}}\)
Tự kết luận tập nghiệm bạn nhé!

 x²+2x-3 = (x+1)(x-3)

vậy MSC = 2(X+1(X-3) qui đồng mẫu số r làm dc r, đk x khác 1; -3

\(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)

\(\Leftrightarrow\frac{9\left(x-3\right)}{99}+\frac{33\left(x+1\right)}{99}=\frac{11\left(x+7\right)}{99}-\frac{99}{99}\)

\(\Leftrightarrow\frac{9\left(x-3\right)+33\left(x+1\right)}{99}=\frac{11\left(x+7\right)-99}{99}\)

\(\Leftrightarrow9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-99\)

\(\Leftrightarrow9x-27+33x+33=11x+77-99\)

\(\Leftrightarrow42x+6=11x-22\Leftrightarrow42x-11x=-6-22\)

\(\Leftrightarrow31x=-28\Leftrightarrow x=-\frac{28}{31}\)

Vậy phương trình có tập nghiệm S={-28/31}