Đáp án:
Giải thích các bước giải:
$1.25x^2-9=0$
$(=)(5x-3)(5x+3)=0$
\(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{-3}{5}\end{array} \right.\)
$2.(x-3)^2-4=0$
$(=)(x-3-2)(x-3+2)=0$
$(=)(x-5)(x-1)=0$
\(\left[ \begin{array}{l}x-5=0\\x-1=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=5\\x=1\end{array} \right.\)
$3.x^2-2x=24$
$(=)x^2-2x-24=0$
$(=)(x-6)(x+4)=0$
\(\left[ \begin{array}{l}x+6=0\\x-4=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=-6\\x=4\end{array} \right.\)
$5.3(x-1)^2-3x(x-5)=0$
$(=)3x^2-6x+3-3x^2+15x=0$
$(=)9x+3=0$
$(=)9x=-3$
$=)x=\dfrac{-1}{3}$
$6.(6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0$
$(=)(6x-2)^2-2(6x-2)(5x-2)+(5x-2)^2=0$
$(=)(6x-2-5x+2)^2=0$
$(=)x=0$
$7.(x-2)^3-x^2(x-6)=0$
$(=)x^3-6x^2+12x-8-x^3+6x^2=0$
$(=)12x=8$
$(=)x=\dfrac{2}{3}$
$8.(x-1)(x^2+x+1)-x(x+2)(x-2)=0$
$(=)x^3-1-x(x^2-4)=0$
$(=)x^3-1-x^3+4x=0$
$(=)4x-1=0$
$(=)x=\dfrac{1}{4}$
Bài 2:
$1.x^2+5x+7$
$=x^2+4x+4+x+3$
$=(x+2)^2+x+3$
$\text{min$=x+3<=>x+2=0<=>x=-2$}$
$2.x^2-20x+101$
$(=)(x-10)^2+1$
$\text{min$=1<=>x-10=0<=>x=10$}$
$3.4a^2+4a+2$
$=(2a+1)^2+1$
$\text{min$=1<=>2a+1=0<=>a=\dfrac{-1}{2}$}$
4.Không làm được
$5.x^2+3x+7$
$=x^2+2x+1+x+6$
$=(x+1)^2+x+6$
$\text{min$=x+6<=>x+1=0<=>x=-1$}$
