Đáp án:
Giải thích các bước giải:
Bài 1:
$1. 25x^2-9=0$
$⇔(5x-3)(5x+3)=0$
⇔\(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{array} \right.\)
$2.(x-3)^2-4=0$
$⇔(x-3-2)(x-3+2)=0$
$⇔(x-5)(x-1)=0$
⇔\(\left[ \begin{array}{l}x-5=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=1\end{array} \right.\)
$3.\dfrac{25}{16}x^2=\dfrac{81}{49}$
$⇔\dfrac{25}{16}x^2-\dfrac{81}{49}=0$
$⇔\left (\dfrac{5}{4}x-\dfrac{9}{7} \right )\left (\dfrac{5}{4}x+\dfrac{9}{7} \right )=0$
⇔\(\left[ \begin{array}{l}\left (\dfrac{5}{4}x-\dfrac{9}{7} \right )=0\\\left (\dfrac{5}{4}x+\dfrac{9}{7} \right )=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\dfrac{5}{4}x=\dfrac{9}{7}\\\dfrac{5}{4}x=-\dfrac{9}{7}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{36}{35}\\x=-\dfrac{36}{35}\end{array} \right.\)
$4.\left (\dfrac{1}{2}x+3 \right )^2=\dfrac{1}{4}$
$⇔\left (\dfrac{1}{2}x+3-\dfrac{1}{2} \right )\left (\dfrac{1}{2}x+3+\dfrac{1}{2} \right )=0$
$⇔\left (\dfrac{1}{2}x+\dfrac{5}{2} \right )\left (\dfrac{1}{2}x+\dfrac{7}{2} \right )=0$
⇔\(\left[ \begin{array}{l}\dfrac{1}{2}x+\dfrac{5}{2}=0\\\dfrac{1}{2}x+\dfrac{7}{2}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-5\\x=-7\end{array} \right.\)
$5.x^2-2x=24$
$⇔x^2-2x-24=0$
$⇔x^2+4x-6x-24=0$
$⇔(x-6)(x+4)=0$
⇔\(\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\)
$6.x^2+6x=-5$
$⇔x^2+6x+5=0$
$⇔(x+5)(x+1)=0$
⇔\(\left[ \begin{array}{l}x+5=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-5\\x=-1\end{array} \right.\)
$7.(x+4)^2-(x+1)(x-1)=16$
$⇔x^2+8x+16-x^2+1=16$
$⇔8x=-1$
$⇔x=-\dfrac{1}{8}$
$8.(x-3)^2-(x-5)(x+5)=7$
$⇔x-6x+9-x^2+25=7$
$⇔-6x=-27$
$⇔x=\dfrac{9}{2}$
$9.3(x-1)^2-3x(x-5)=1$
$⇔3x^2-6x+3-3x^2+15x=1$
$⇔9x=-2$
$⇒x=-\dfrac{2}{9}$
$10.(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0$
$⇔4x^2-4x+1+x^2+6x+9-5x^2+245=0$
$⇔2x=-254$
$⇒x=-127$
$11.(x+3)^2+(x-2)^2=2x^2$
$⇔x^2+6x+9+x^2-4x+4-2x^2=0$
$⇔2x=-13$
$⇔x=-\dfrac{13}{2}$
$12.8x^3-12x^2+6x-1=0$
$⇔(2x-1)^3=0$
$⇔2x-1=0$
$⇔x=\dfrac{1}{2}$
$13.(2x+1)(4x^2-2x+1)-8x(x^2+2)=17$
$⇔8x^3+1-8x^3-16x=17$
$⇔-16x=16$
$⇔x=-1$
$14.4x^2-12x+9=0$
$⇔(2x-3)^2=0$
$⇔2x-3=0$
$⇔x=\dfrac{3}{2}$
$15.9x^2-30x+24=0$
$⇔9x^2-18x-12x+24=0$
$⇔9x(x-2)-12(x-2)=0$
$⇔(9x-12)(x-2)=0$
⇔\(\left[ \begin{array}{l}9x-12=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=2\end{array} \right.\)
$16.(3x-2)(3x+2)-9(x-1)x=0$
$⇔9x^2-4-9x^2+9x=0$
$⇔9x=4$
$⇔x=\dfrac{4}{9}$
