Toán Lớp 8: Giả phương trình:
a, (7x-3)/(x-1)=2/3
b, (5x-1)/(3x+2)=(5x-7)/(3x-1)
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Toán Lớp 8: Giả phương trình:
a, (7x-3)/(x-1)=2/3
b, (5x-1)/(3x+2)=(5x-7)/(3x-1)
Comments ( 2 )
a,
(7x-3)/(x-1)=2/3 ĐKXĐ : x\ne1
⇔2.(x-1)=3.(7x-3)
⇔2x-2=21x-9
⇔2x-21x=-9+2
⇔-19x=-7
⇔x=7/19 (TMĐK)
Vậy tập nghiệm của phương trình là S={7/19}
b,
(5x-1)/(3x+2)=(5x-7)/(3x-1) ĐKXĐ : x\ne-2/3 , x\ne1/3
⇔(5x-1)(3x-1)=(5x-7)(3x+2)
⇔15x^2-5x-3x+1=15x^2+10x-21x-14
⇔(15x^2-15x^2)+(-5x-3x-10x+21x)=-14-1
⇔3x=-15
⇔x=-5 (TMĐK)
Vậy tập nghiệm của phương trình là S={-5}
#andy
\[\begin{array}{l}
a)\dfrac{{7x – 3}}{{x – 1}} = \dfrac{2}{3}\\
DKXD:x – 1 \ne 0\\
\Leftrightarrow x \ne 1\\
\Leftrightarrow \dfrac{{3\left( {7x – 3} \right)}}{{x\left( {x – 1} \right)}} = \dfrac{{2\left( {x – 1} \right)}}{{x\left( {x – 1} \right)}}\\
\Leftrightarrow 3\left( {7x – 3} \right) = 2\left( {x – 1} \right)\\
\Leftrightarrow 21x – 9 = 2x – 2\\
\Leftrightarrow 21x – 2x = – 2 + 9\\
\Leftrightarrow 19x = 7\\
\Leftrightarrow x = \dfrac{7}{{19}}\left( {TMĐK} \right)\\
b)\dfrac{{5x – 1}}{{3x + 2}} = \dfrac{{5x – 7}}{{3x – 1}}\\
DKXD:\left\{ \begin{array}{l}
3x + 2 \ne 0\\
3x – 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne – \dfrac{2}{3}\\
x \ne \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left( {3x – 1} \right)\left( {5x – 1} \right) = \left( {3x + 2} \right)\left( {5x – 7} \right)\\
\Leftrightarrow 15{x^2} – 3x – 5x + 1 = 15{x^2} – 21x + 10x – 14\\
\Leftrightarrow 15{x^2} – 3x – 5x – 15{x^2} + 21x – 10x = – 14 – 1\\
\Leftrightarrow 3x = – 15\\
\Leftrightarrow x = – 5\left( {TMĐK} \right)
\end{array}\]