Đáp án:
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Giải thích các bước giải:
$57/ x(x+3)(x-3) - (x+2)(x^2-2x+4)=0$
$ ⇔x(x^2-9)-(x^3-8)=0$
$⇔x^3 -9x -x^3 +8=0$
$⇔x^3 -x^3- 9x +8=0$
$⇔-9x = -8$
$⇔x= \dfrac{8}{9}$
Vậy....
$58) 2x(x-3)+5(x-3)=0$
$⇔(x-3)(2x+5)=0$
⇔\(\left[ \begin{array}{l}x-3=0\\2x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-\dfrac{5}{2}\end{array} \right.\)
Vậy....
$59/ (3x-1)(x^2+2)= (3x-1)(7x-10)$
$ ⇔(3x-1)(x^2+2) -(3x-1)(7x-10)=0$
$⇔(3x-1)(x^2-7x+12)=0$
⇔\(\left[ \begin{array}{l}3x-1=0\\x^2-7x+12=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=4\\x=3\end{array} \right.\)
Vậy....
$60/ (x+2)(3-4x)=x^2+4x+4$
$⇔(x+2)(3-4x)-(x^2+4x+4)=0$
$⇔(x+2)(3-4x)-(x+2)² =0$
$⇔(x+2)(3-4x-x-2)=0$
$⇔(x+2)(-5x+1)=0$
⇔\(\left[ \begin{array}{l}x+2=0\\-5x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy,..
$61/ x(2x-7)-4x+14=0$
$⇔ x(2x-7)-2(2x-7)=0$
$⇔(2x-7)(x-2)=0$
⇔\(\left[ \begin{array}{l}2x-7=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=2\end{array} \right.\)
Vậy..
$62/ 3x-15 =2x(x-5)$
$⇔3(x-5)-2x(x-5)=0$
$⇔(x-5)(3-2x)=0$
⇔\(\left[ \begin{array}{l}x-5=0\\3-2x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy...
$63/(2x+1)(3x-2)=(5x-8)(2x+1)$
$⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0$
$⇔(2x+1)(3x-2-5x+8)=0$
$⇔(2x+1)(-2x+6)=0$
⇔\(\left[ \begin{array}{l}2x+1=0\\-2x+6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=3\end{array} \right.\)
Vậy..
$64/0,5x(x-3)=(x-3)(1,5x-1)$
$⇔ 0,5x(x-3)-(x-3)(1,5x-1)=0$
$⇔(x-3)(0,5x-1,5x+1)=0$
$⇔(x-3)(-x+1)=0$
⇔\(\left[ \begin{array}{l}x-3=0\\-x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
Vậy...
$65/ (2x^2+1)(4x-3) = (x-12)(2x^2+1)$
$⇔(2x^2+1)(4x-3) -(x-12)(2x^2+1)=0$
Do $2x^2+1 \neq 0$ (loại)
$⇔(4x-3-x+12)=0$
$⇔3x+9=0$
$⇔3x= -9$
$⇔x = -9 : 3$
$⇔x = -3$
Vậy...
$66/ (x-1)(5x+3)=(3x-8)(x-1)$
$⇔(x-1)(5x+3)-(3x-8)(x-1)=0$
$⇔(x-1)(5x+3-3x+8)=0$
$⇔(x-1)(2x+11)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-\dfrac{11}{2}\end{array} \right.\)
Vậy....
$67/ (2-3x)(x+11)=(3x-2)(2-5x)$
$⇔-(3x-2)(x+11)-(3x-2)(2-5x)=0$
$⇔(3x-2)(-x-11-2+5x)=0$
$⇔(3x-2)(4x-13)=0$
⇔\(\left[ \begin{array}{l}3x-2=0\\4x-13=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{array} \right.\)
Vậy...
$71/ (x+2)(x-3)(17x^2-17x+8)=(x+2)(x-3)(x^2-17x+33)$
$⇔(x+2)(x-3)(17x^2 -17x+8)-(x+2)(x-3)(x^2-17x+33)$
$⇔(x+2)(x-3)(17x^2-17x+8-x^2+17x-33)=0$
$⇔(x+2)(x-3)(16x^2-25)=0$
⇔\(\left[ \begin{array}{l}x+2=0\\x-3=0\\16x^2-25=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=3\\x=±\dfrac{5}{4}\end{array} \right.\)
Vậy....
