T
thaosida
Câu 18:
ĐK: $ x \leq1-\sqrt{2}\bigcup x \geq 1+\sqrt{2}$
$2\sqrt{x^2-2x+1}+\sqrt[3]{x^3-14}=x-2$
Đặt x=a+1
PT $\Rightarrow \sqrt{(a+1)^2-2(a+1)-1}+ \sqrt[3]{(a+1)^3-14}=a - 1$
$\Leftrightarrow 2\sqrt{a^2-2}+ \sqrt[3]{a^3+3^2-13}-(a-1)=0$
$\Leftrightarrow 2\sqrt{a^2-2}+ \frac{6(a^2-2a)}{A}=0$
$\Rightarrow \sqrt{a^2-2}= 0$
$\Rightarrow(x-1)^2-2=0$
$\Leftrightarrow x=1-\sqrt{2}\bigcup x=1+\sqrt{2}$
ĐK: $ x \leq1-\sqrt{2}\bigcup x \geq 1+\sqrt{2}$
$2\sqrt{x^2-2x+1}+\sqrt[3]{x^3-14}=x-2$
Đặt x=a+1
PT $\Rightarrow \sqrt{(a+1)^2-2(a+1)-1}+ \sqrt[3]{(a+1)^3-14}=a - 1$
$\Leftrightarrow 2\sqrt{a^2-2}+ \sqrt[3]{a^3+3^2-13}-(a-1)=0$
$\Leftrightarrow 2\sqrt{a^2-2}+ \frac{6(a^2-2a)}{A}=0$
$\Rightarrow \sqrt{a^2-2}= 0$
$\Rightarrow(x-1)^2-2=0$
$\Leftrightarrow x=1-\sqrt{2}\bigcup x=1+\sqrt{2}$
Last edited by a moderator: