x=0 ko là nghiệm
chia cả hai vê cho x<>0, ta được:
\(x-\dfrac{1}{x}+\sqrt[3]{x-\dfrac{1}{x}}=2\)
Đặt \(\sqrt[3]{x-\dfrac{1}{x}}=a\)
=>a^3+a=2
=>a=1
=>x-1/x=1
=>\(x=\dfrac{1\pm\sqrt{5}}{2}\)
GPT :
\(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
GPT: \(\sqrt{2x+3-\sqrt{x+2}}+\sqrt{2x+4+\sqrt{x+2}}=1+2\sqrt{x+2}\)
GPT :
\(\sqrt[4]{2x^3+x^2-2x+8}+\sqrt[4]{3x^3+x^2-2x+10}=2\sqrt[4]{x^2-2x+4}\)
gpt:
\(x^4-2x^3+x=\sqrt{(x^2-x).2}\)
\(x(5x^3+2)-2(\sqrt{2x+1}-1)=0\)
\(\sqrt{x+\frac{3}{x}}=\frac{x^2-7}{2x+2}\)
GPT : \(\sqrt{\left|x+1\right|}+\sqrt[4]{x^2-2x+5}=2\sqrt[4]{x^2+3}\)
gpt:
\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)
Gpt: \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
GPT
\(\sqrt[]{2x+3+\sqrt{x+2}}+\sqrt[]{2x+2-\sqrt{2+x}}=1+2\sqrt{x+2}\)
GPT:\(\sqrt{1-x}+\sqrt{4+x}=3...\)
\(x^2+4x+5=2\sqrt{2x+3}\)
