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\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(-537x^2+5054x=-541x^2+5092x\)
\(-537x^2+5054x+541x^2-5092x=0\)
\(4x^2-38x=0\)
\(x\left(2x-19\right)=0\)
\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)
c) ĐK: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2(x-11)}{x^2-4}\)
\(\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{x^2-7x-2}{(x-2)(x+2)}=\frac{2x-22}{(x-2)(x+2)}\)
\(\Rightarrow x^2-7x-2=2x-22\)
\(\Leftrightarrow x^2-9x+20=0\Leftrightarrow (x-4)(x-5)=0\Rightarrow x=4\) hoặc $x=5$
(đều thỏa mãn)
d) ĐK: \(x^2-6x+7\neq 0\)
PT \(\Leftrightarrow (x^2-6x+7)+\frac{14}{x^2-6x+7}-9=0\)
\(\Rightarrow (x^2-6x+7)^2-9(x^2-6x+7)+14=0\)
\(\Leftrightarrow (x^2-6x+7-2)(x^2-6x+7-7)=0\)
\(\Leftrightarrow (x^2-6x+5)(x^2-6x)=0\)
\(\Leftrightarrow (x-1)(x-5)x(x-6)=0\)
\(\Rightarrow x\in \left\{1;5;0;6\right\}\) (đều thỏa mãn)
Vậy.........
a) ĐKXĐ: $x\neq 1$
PT \(\Leftrightarrow \frac{x^2+x+1+2(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow \frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)
\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow 2x^2-3x+1=0\)
\(\Leftrightarrow (x-1)(2x-1)=0\)
Mà $x\neq 1$ nên $2x-1=0\Rightarrow x=\frac{1}{2}$ là nghiệm
b) ĐK: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3-x}{2-x}=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{3-x}{2-x}=\frac{6-x}{3(x^2-4)}\)
\(\Leftrightarrow \frac{1}{x+2}+\frac{3-x}{x-2}=\frac{6-x}{3(x-2)(x+2)}\)
\(\Leftrightarrow \frac{-x^2+2x+4}{(x-2)(x+2)}=\frac{6-x}{3(x-2)(x+2)}\)
\(\Rightarrow 3(-x^2+2x+4)=6-x\)
\(\Leftrightarrow -3x^2+7x+6=0\)
\(\Leftrightarrow (x-3)(3x+2)=0\Rightarrow x=3\) hoặc $x=-\frac{2}{3}$
Vậy........
a) \(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(\Leftrightarrow x^2+x-2x-2=x^2-4\)
\(\Leftrightarrow x^2-x-2=x^2-4\)
\(\Leftrightarrow-x-2=-4\)
\(\Leftrightarrow-x=-4+2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)
Vậy: phương trình có tập nghiệm: S = {2}
a) \(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x+1=x+2\)
\(\Leftrightarrow x+1-x-2=0\)
\(\Leftrightarrow-1=0\left(vl\right)\)
Vậy pt vô no
b) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{2x-4-x-1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{-5}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow-5\left(x+2\right)=\left(3x-11\right)\left(x-2\right)\)
\(-5x+2=3x^2-11x-6x+22\)
\(3x^2-17x+22+5x-2=3x^2-12x+20=0\)
đến đây mk chịu ~
Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(\frac{2}{x+1}+\frac{1}{2-x}=\frac{3x-11}{x^2-x-2}\)
\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)-\left(x+1\right)-\left(3x-11\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow2x-4-x-1-3x+11=0\)
\(\Leftrightarrow-2x+6=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là \(S=\left\{3\right\}\)