Bài 27: ĐKXĐ: x>=-2
Ta có: \(3\sqrt[3]{x+6}-\sqrt{x+2}=4\)
=>\(3\cdot\sqrt[3]{x+6}-6+2-\sqrt{x+2}=4-6+2=0\)
=>\(3\left(\sqrt[3]{x+6}-2\right)+\frac{4-x-2}{\sqrt{x+2}+2}=0\)
=>\(3\cdot\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{x-2}{\sqrt{x+2}+2}=0\)
=>\(3\cdot\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{x-2}{\sqrt{x+2}+2}=0\)
=>\(\left(x-2\right)\left(\frac{3}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{1}{\sqrt{x+2}+2}\right)=0\)
=>x-2=0
=>x=2(nhận)
Bài 25:
ĐKXĐ: \(x^3+3x\ge0\)
=>\(x\left(x^2+3\right)\ge0\)
=>x>=0
Ta có: \(\left(x+6\right)^2-8\sqrt{x^3+3x}=33\)
=>\(\left(x+6\right)^2-81-8\sqrt{x^3+3x}+48=33+48-81=0\)
=>\(\left(x+6-9\right)\left(x+6+9\right)-8\left(\sqrt{x^3+3x}-6\right)=0\)
=>\(\left(x-3\right)\left(x+15\right)-8\cdot\frac{x^3+3x-36}{\sqrt[2]{x^2+3x}+6}=0\)
=>\(\left(x-3\right)\left(x+15\right)-8\cdot\frac{x^3-3x^2+3x^2-9x+12x-36}{\sqrt[2]{x^2+3x}+6}=0\)
=>\(\left(x-3\right)\left(x+15\right)-8\cdot\frac{\left(x-3\right)\left(x^2+3x+12\right)}{\sqrt[2]{x^2+3x}+6}=0\)
=>\(\left(x-3\right)\left\lbrack x+15-\frac{8\left(x^2+3x+12\right)}{\sqrt{x^3+3x}+6}\right\rbrack=0\)
=>x-3=0
=>x=3(nhận)
