Toán Lớp 8: Giải pt sau :
1. $\dfrac{(x-2)(x+10)}{3}$ – $\dfrac{(x+4) (x+10) }{12}$ = $\dfrac{(x-2) (x+4) }{4}$
2. $\dfrac{(2x-3)(2x+3) }{8}$ = $\dfrac{(x-4)^2 }{6}$ + $\dfrac{(x-2)^2 }{3}$
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Toán Lớp 8: Giải pt sau :
1. $\dfrac{(x-2)(x+10)}{3}$ – $\dfrac{(x+4) (x+10) }{12}$ = $\dfrac{(x-2) (x+4) }{4}$
2. $\dfrac{(2x-3)(2x+3) }{8}$ = $\dfrac{(x-4)^2 }{6}$ + $\dfrac{(x-2)^2 }{3}$
Comments ( 1 )
Giải đáp:
\(\begin{array}{l}
1)x = 8\\
2)x = \dfrac{{123}}{{64}}
\end{array}\)
Lời giải và giải thích chi tiết:
\(\begin{array}{l}
1)\dfrac{{{x^2} + 8x – 20}}{3} – \dfrac{{{x^2} + 14x + 40}}{{12}} = \dfrac{{{x^2} + 2x – 8}}{4}\\
\to 4\left( {{x^2} + 8x – 20} \right) – {x^2} – 14x – 40 – 3\left( {{x^2} + 2x – 8} \right) = 0\\
\to 4{x^2} + 32x – 80 – {x^2} – 14x – 40 – 3{x^2} – 6x + 24 = 0\\
\to 12x = 96\\
\to x = 8\\
2)\dfrac{{4{x^2} – 9}}{8} = \dfrac{{{x^2} – 8x + 16}}{6} + \dfrac{{{x^2} – 4x + 4}}{3}\\
\to 3\left( {4{x^2} – 9} \right) = 4\left( {{x^2} – 8x + 16} \right) + 8\left( {{x^2} – 4x + 4} \right)\\
\to 12{x^2} – 27 = 4{x^2} – 32x + 64 + 8{x^2} – 32x + 32\\
\to 64x = 123\\
\to x = \dfrac{{123}}{{64}}
\end{array}\)