5: Đặt \(\sqrt[3]{2x-5}=a;\sqrt[3]{3x+7}=b\)
=>\(a^3+b^3=2x-5+3x+7=5x+2\)
Ta có: \(\sqrt[3]{2x-5}+\sqrt[3]{3x+7}=\sqrt[3]{5x+2}\)
=>\(a+b=\sqrt[3]{a^3+b^3}\)
=>\(a^3+b^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
=>3ab(a+b)=0
=>ab(a+b)=0
TH1: a=0
=>2x-5=0
=>2x=5
=>x=2,5
TH2: 3x+7=0
=>3x=-7
=>\(x=-\frac73\)
TH3: a+b=0
=>a=-b
=>\(\sqrt[3]{2x-5}=-\sqrt[3]{3x+7}\)
=>3x+7=-2x+5
=>3x+2x=5-7
=>5x=-2
=>\(x=-\frac25\)
8:
ĐKXĐ: x>=-3/2
\(\sqrt{x^2+1}+\sqrt{x^2+2x+4}=\sqrt{9+6x}\)
=>\(\sqrt{x^2+1}-1+\sqrt{x^2+2x+4}-2=\sqrt{9+6x}-3\)
=>\(\frac{x^2+1-1}{\sqrt{x^2+1}+1}+\frac{x^2+2x+4-4}{\sqrt{x^2+2x+4}+2}=\frac{9+6x-9}{\sqrt{9+6x}+3}\)
=>\(\frac{x^2}{\sqrt{x^2+1}+1}+\frac{x^2+2x}{\sqrt{x^2+2x+4}+2}=\frac{6x}{\sqrt{9+6x}+3}\)
=>\(x\left(\frac{x}{\sqrt{x^2+1}+1}+\frac{x+2}{\sqrt{x^2+2x+4}+2}-\frac{6}{\sqrt{9+6x}+3}\right)=0\)
=>x=0(nhận)
