Đáp án:
Giải thích các bước giải:
\(1/\\ a,\ 4x^{2}-12x+9\\ =(2x)^{2}-2.2x.3+3^{2}\\ =(2x-3)^{2}\\ b,\ (x-7)^{2}-81\\ =(x-7)^{2}-9^{2}\\ =(x-7-9)(x-7+9)\\ =(x-16)(x+2)\\ c,\ (x-5)^{2}=x-5\\ ⇔(x-5)^{2}-(x-5)=0\\ ⇔(x-5)(x-5-1)=0\\ ⇔(x-5)(x-6)=0\\ ⇔\left[ \begin{array}{l}x-5=0\\x-6=0\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=5\\x=6\end{array} \right.\\ \text{Vậy x = 5 hoặc x = 6}\\ d,\ x^{2}-8x+16=0\\ ⇔x^{2}-2.x.4+4^{2}=0\\ ⇔(x-4)^{2}=0\\ ⇔x-4=0\\ ⇔x=4\\ \text{Vậy x = 4}\\ e,\ 8x^{3}+\dfrac{1}{27}\\ =(2x)^{3}+\bigg(\dfrac{1}{3}\bigg)^{3}\\ =\bigg(2x+\dfrac{1}{3}\bigg)\bigg(4x^{2}-\dfrac{2}{3}x+\dfrac{1}{9}\bigg)\\ g,\ 125-x^{6}\\ =5^{3}-(x^{2})^{3}\\ =(5-x^{2})(25+5x^{2}+x^{4})\\ h,\ x^{2}-2x-4y^{2}+1\\ =(x^{2}-2x+1)-(2y)^{2}\\ =(x-1)^{2}-(2y)^{2}\\ =(x-1-2y)(x-1+2y)\\ 2/\\ a,\ (x-8)^{2}-49=0\\ ⇔(x-8)^{2}-7^{2}=0\\ ⇔(x-8-7)(x-8+7)=0\\ ⇔(x-15)(x-1)=0\\ ⇔\left[ \begin{array}{l}x-15=0\\x-1=0\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=15\\x=1\end{array} \right.\\ \text{Vậy x = 15 hoặc x = 1}\\ b,\ x^{2}+10x+25=0\\ ⇔x^{2}+2.x.5+5^{2}=0\\ ⇔(x+5)^2=0\\ ⇔x+5=0\\ ⇔x=(-5)\\ \text{Vậy x = (- 5)}\\ c,\ 4x^{2}-8x=-4\\ ⇔4x^{2}-8x+4=0\\ ⇔4(x^{2}-2x+1)=0\\ ⇔(x-1)^{2}=0\\ ⇔x-1=0\\ ⇔x=1\\ \text{Vậy x = 1}\\ d,\ (x+11)^{2}=x+11\\ ⇔(x+11)^2-(x+11)=0\\ ⇔(x+11)(x+11-1)=0\\ ⇔(x+11)(x+10)=0\\ ⇔\left[ \begin{array}{l}x+11=0\\x+10=0\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=-11\\x=-10\end{array} \right.\\ \text{Vậy x = (- 11) hoặc x = (- 10)}\\ e,\ x^{3}-16x=0\\ ⇔x(x^{2}-16)=0\\ ⇔x(x-4)(x+4)=0\\ ⇔\left[ \begin{array}{l}x=0\\x-4=0\\x+4=0\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=0\\x=4\\x=-4\end{array} \right.\\ \text{Vậy x $\in$ {0; 4; -4}}\\ g,\ x^{4}-2x^{3}+10x^{2}-20x=0\\ ⇔x^{3}(x-2)+10x(x-2)=0\\ ⇔(x^{3}+10x)(x-2)=0\\ ⇔x(x^{2}+10)(x-2)=0\\ ⇔\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\ \text{(Vì $x^2+10>0$ ∀ x)}\\ ⇔\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\\ \text{Vậy x = 0 hoặc x = 2}\)
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