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 ....

\(DKXD:x>0\)

\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)

\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)

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

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

\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)

\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)

\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy PT có 2 nghiệm \(x=1;x=3\)

tai sao cho xcan bac hai lai bang 2

tai sao  \(x\sqrt{x+\frac{3}{x}}=2\) ?

gõ lại đề 

\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)

Điều kiện: \(x\ge\frac{1}{3}\)

Đặt \(\sqrt{x-\frac{1}{3}}=a\left(a\ge0\right)\)

\(\Rightarrow x=a^2+\frac{1}{3}\)

Ta suy ra phương trình tương đương với

\(18\left(a^2+\frac{1}{3}\right)^2-2\left(a^2+\frac{1}{3}\right)-\frac{17}{3}+9a=0\)

\(\Leftrightarrow54a^4+30a^2+27a-13=0\)

\(\Leftrightarrow\left(3a-1\right)\left(18a^3+6a^2+12a+13\right)=0\)

Dễ thấy \(18a^3+6a^2+12a+13>0\) vì \(a\ge0\)

\(\Rightarrow3a-1=0\)

\(\Leftrightarrow a=\frac{1}{3}\)

\(\Leftrightarrow\sqrt{x-\frac{1}{3}}=\frac{1}{3}\)

\(\Leftrightarrow x-\frac{1}{3}=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{4}{9}\)

b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)

=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)

=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

=> x - 100 = 0

=> x = 100

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)

\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)

\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)

\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)

\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)

Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)

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

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

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

\(\Leftrightarrow\left(2x-5\right).\left(\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x^2-5x+4=x^2-5x+6\left(loai\right)\end{cases}}}\)

Vậy..