Toán Lớp 8: giải các phương trình sau :
a, $\frac{3x-2}{x+7}$ = $\frac{6x+1}{2x-3}$
b, $\frac{x+1}{x-1}$ – $\frac{x-1}{x+1}$ =$\frac{4}{x^{2}-1}$
c, $\frac{6}{x-5}$+$\frac{2}{x-8}$=$\frac{18}{(x-5).(8-x)}$-1
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Toán Lớp 8: giải các phương trình sau :
a, $\frac{3x-2}{x+7}$ = $\frac{6x+1}{2x-3}$
b, $\frac{x+1}{x-1}$ – $\frac{x-1}{x+1}$ =$\frac{4}{x^{2}-1}$
c, $\frac{6}{x-5}$+$\frac{2}{x-8}$=$\frac{18}{(x-5).(8-x)}$-1
Comments ( 1 )
$a)$ $\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}$
$⇔6x^2-13x+6=6x^2+43x+7$
$⇔6x^2-13x+6-6=6x^2+43x+7-6$
$⇔6x^2-13x=6x^2+43x+1$
$⇔6x^2-13x-\left(6x^2+43x\right)=6x^2+43x+1-\left(6x^2+43x\right)$
$⇔-56x=1$
$⇔x=-\dfrac{1}{56}$
$\text{Vậy pt có nghiệm S ={$-\dfrac{1}{56}$}}$
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$b)$ $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}$
$⇔4x=4$
$⇔x=1$
$\text{Vậy pt vô nghiệm.}$
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$c)$ $\dfrac{6}{x-5}+\dfrac{2}{x-8}=\dfrac{18}{\left(x-5\right)\cdot \left(8-x\right)}-1$
$⇔6\left(x-8\right)\left(-x+8\right)+2\left(x-5\right)\left(-x+8\right)=-18\left(-x+8\right)-\left(x-5\right)\left(x-8\right)\left(-x+8\right)$
⇔122x−8x ^2 −162x+21x ^2 −x ^3 =0
⇔−40x+13x^2−x^3=0
$⇔-x\left(x-5\right)\left(x-8\right)=0$
\(⇔\left[ \begin{array}{l}x=0\\\:x-5=0\quad \\x-8=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=0\\\:x=5(loại)\quad \\x=8(loại)\end{array} \right.\)
$\text{Vậy pt có nghiệm S = {0}}$