Toán Lớp 8: Giải các phương trình sau:
a) -$\frac{3}{2}$ x + $\frac{5}{4}$ = $\frac{5}{6}$ x + 2
b) $\frac{10x+5}{12}$ = 1 + $\frac{3+7x}{9}$
c) $(x+1)^{2}$ – 3(x+2) = (x-1)(x+1)
d) 5x(x-3) – (x-2)(x+2) = $(2x+1)^{2}$
Làm ý nào trước cũng được ạ ????♂️
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$a)$ $-\dfrac{3}{2}x+\dfrac{5}{4}=\dfrac{5}{6}x+2$
$⇔-\dfrac{3}{2}x+\dfrac{5}{4}-\dfrac{5}{4}=\dfrac{5}{6}x+2-\dfrac{5}{4}$
$⇔-\dfrac{3}{2}x=\dfrac{3}{4}+\dfrac{5}{6}x$
$⇔-\dfrac{3}{2}x-\dfrac{5}{6}x=\dfrac{3}{4}+\dfrac{5}{6}x-\dfrac{5}{6}x$
$⇔-\dfrac{7}{3}x=\dfrac{3}{4}$
$⇔3\left(-\dfrac{7}{3}x\right)=\dfrac{3\cdot \:3}{4}$
$⇔-7x=\dfrac{9}{4}$
$⇔\dfrac{-7x}{-7}=\dfrac{\frac{9}{4}}{-7}$
$⇔x=-\dfrac{9}{28}$
Vậy pt có nghiệm S = {-\frac{9}{28}}
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$b)$ $\dfrac{10x+5}{12}=1+\dfrac{3+7x}{9}$
$⇔\dfrac{10x+5}{12}\cdot \:36=1\cdot \:36+\dfrac{3+7x}{9}\cdot \:36$
$⇔3\left(10x+5\right)=36+4\left(7x+3\right)$
$⇔30x+15=28x+48$
$⇔30x+15-15=28x+48-15$
$⇔30x=28x+33$
$⇔30x-28x=28x+33-28x$
$⇔2x=33$
$⇔\dfrac{2x}{2}=\dfrac{33}{2}$
$⇔x=\dfrac{33}{2}$
Vậy pt có nghiệm S = {\frac{33}{2}}
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$c)$ $\left(x+1\right)^{2\:}\:-\:3\left(x+2\right)\:=\:\left(x-1\right)\left(x+1\right)$
$⇔x^2-x-5=x^2-1$
$⇔x^2-x-5+5=x^2-1+5$
$⇔x^2-x=x^2+4$
$⇔x^2-x-x^2=x^2+4-x^2$
$⇔-x=4$
$⇔\dfrac{-x}{-1}=\dfrac{4}{-1}$
$⇔x=-4$
Vậy pt có nghiệm S = {-4}
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$d)$ $5x\left(x-3\right)\:-\:\left(x-2\right)\left(x+2\right)\:=\:\left(2x+1\right)^{2\:}$
$⇔4x^2-15x+4=4x^2+4x+1$
$⇔4x^2-15x+4-4=4x^2+4x+1-4$
$⇔4x^2-15x=4x^2+4x-3$
$⇔4x^2-15x-\left(4x^2+4x\right)=4x^2+4x-3-\left(4x^2+4x\right)$
$⇔-19x=-3$
$⇔\dfrac{-19x}{-19}=\dfrac{-3}{-19}$
$⇔x=\dfrac{3}{19}$
Vậy pt có nghiệm S = {\frac{3}{19}}