6: Ta có: \(\sqrt[3]{1-x}+\sqrt[3]{1+x}=2\)
=>\(\sqrt[3]{1-x}-1+\sqrt[3]{1+x}-1=2-2=0\)
=>\(\frac{1-x-1}{\sqrt[3]{\left(1-x\right)^2}+\sqrt[3]{1-x}+1}+\frac{1+x-1}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{1+x}+1}=0\)
=>\(\frac{-x}{\sqrt[3]{\left(1-x\right)^2}+\sqrt[3]{1-x}+1}+\frac{x}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{1+x}+1}=0\)
=>\(x\left(\frac{-1}{\sqrt[3]{\left(1-x\right)^2}+\sqrt[3]{1-x}+1}+\frac{1}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{1+x}+1}\right)=0\)
=>x=0
7: ĐKXĐ: 3-2x>=0
=>2x<=3
=>x<=3/2
Ta có: \(\sqrt[3]{5+3x}+\sqrt{3-2x}=3\)
=>\(\sqrt[3]{3x+5}-2+\sqrt{3-2x}-1=0\)
=>\(\frac{3x+5-8}{\sqrt[3]{\left(3x+5\right)^2}+2\cdot\sqrt[3]{3x+5}+4}+\frac{3-2x-1}{\sqrt{3-2x}+1}=0\)
=>\(\frac{3x-3}{\sqrt[3]{\left(3x+5\right)^2}+2\cdot\sqrt[3]{3x+5}+4}+\frac{-2x+2}{\sqrt{3-2x}+1}=0\)
=>\(\) \(\left(x-1\right)\left(\frac{3}{\sqrt[3]{\left(3x+5\right)^2}+2\cdot\sqrt[3]{3x+5}+4}+\frac{-2}{\sqrt{3-2x}+1}\right)=0\)
=>x-1=0
=>x=1(nhận)
9: ĐKXĐ: -1<=x<=1
Đặt \(\sqrt{1-x}=a;\sqrt{1+x}=b\)
\(a\cdot b=\sqrt{1-x}\cdot\sqrt{1+x}=\sqrt{1-x^2}\)
Ta có: \(2-2x^2+4\cdot\sqrt{1-x^2}=\left(\sqrt{1-x}+\sqrt{1+x}\right)\left(2+\sqrt{1-x^2}\right)\)
=>\(2\left(1-x^2\right)+4\cdot\sqrt{1-x^2}=\left(\sqrt{1-x}+\sqrt{1+x}\right)\left(2+\sqrt{1-x^2}\right)\)
=>\(2a^2b^2+4ab=\left(a+b\right)\left(2+ab\right)\)
=>\(2ab\left(ab+2\right)-\left(a+b\right)\left(ab+2\right)=0\)
=>(ab+2)(2ab-a-b)=0
=>2ab-a-b=0
=>2ab=a+b
=>\(2\sqrt{1-x^2}=\sqrt{1-x}+\sqrt{1+x}\)
=>\(2\left(\sqrt{1-x^2}-1\right)=\sqrt{1-x}-1+\sqrt{1+x}-1\)
=>\(2\cdot\frac{1-x^2-1}{\sqrt{1-x^2}+1}=\frac{1-x-1}{\sqrt{1-x}+1}+\frac{1+x-1}{\sqrt{1+x}+1}\)
=>\(\frac{-2x^2}{\sqrt{1-x^2}+1}=\frac{-x}{\sqrt{1-x}+1}+\frac{x}{\sqrt{1+x}+1}\)
=>\(x\left(-\frac{2x}{\sqrt{1-x^2}+1}+\frac{1}{\sqrt{1-x}+1}-\frac{1}{\sqrt{1+x}+1}\right)=0\)
=>x=0(nhận)
