Đáp án:
Giải thích các bước giải:
$25x^2-9=0$
$⇔(5x-3)(5x+3)=0$
⇔\(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{array} \right.\)
$(x-3)^2-4=0$
$⇔(x-3-2)(x-3+2)=0$
$⇔(x-5)(x-1)=0$
⇔\(\left[ \begin{array}{l}x=5\\x=1\end{array} \right.\)
$x^2-2x=24$
$⇔x^2-2x-24=0$
$⇔x^2-6x+4x-24=0$
$⇔(x-6)(x+4)=0$
⇔\(\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\)
$(x+4)^2-(x+1)(x-1)=16$
$⇔x^2+8x+16-x^2+1=16$
$⇔8x=16-17$
$⇔8x=-1$
$⇔x=-\dfrac{1}{8}$
$(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0$
$⇔4x^2-4x+1+x^2+6x+9-5(x^2-49)=0$
$⇔5x^2+2x+10-5x^2+245=0$
$⇔2x=-255$
$⇔x=-\dfrac{255}{2}$
$(6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0$
$⇔36x^2-24x+4+25x^2-20x+4-12x(5x-2)+4(5x-2)=0$
$⇔61x^2-44x+8-60x^2+24x+20x-8=0$
$⇔x^2=0$
$⇔x=0$
$3(x-1)^2-3x(x-5)=1$
$⇔3(x^2-2x+1)-3x^2+15x=1$
$⇔3x^2-6x+3-3x^2+15x=1$
$⇔9x=1-3$
$⇔9x=-2$
$⇔x=-\dfrac{2}{9}$
$(x-2)^3-x^2(x-6)=4$
$⇔x^3-6x^2+12x-8-x^3+6x^2=4$
$⇔12x=12$
$⇔x=1$
$(x-1)(x^2+x+1)-x(x+2)(x-2)=5$
$⇔x^3-1-x(x^2-4)=5$
$⇔x^3-1-x^3+4x=5$
$⇔4x=6$
$⇔x=\dfrac{3}{2}$
$(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=0$
$⇔x^3-3x^2+3x-1-x^3-27+3x^2-12=0$
$⇔3x-40=0$
$⇔3x=40$
$⇔x=\dfrac{40}{3}$
