Question

[Nguyễn Bá Minh]

Giải phương trình:

a,\(\left(x+1\right)\sqrt{\frac{1}{x^3+1}}-2\sqrt{\frac{x^2-x+1}{x+1}}+1=0\)

b,\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)

c,\(2-x^2=\sqrt{2-x}\)

d,\(x^3+1=2\sqrt[3]{2x-1}\)

e\(2\left(x^2+x+\frac{1}{2}\right)=\sqrt{4x+1}\)

f,\(\sqrt[3]{2-x}+\sqrt{x-1}=1\)

g,\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)

Answer 1

[Phan Văn Hiếu]

a) đkxđ x>-1

\(\left(x+1\right)\sqrt{\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}}-2\sqrt{\frac{x^2-x+1}{x+1}}+1=0\)

\(=\sqrt{\frac{x+1}{x^2-x+1}}-2\sqrt{\frac{x^2-x+1}{x+1}}+1=0\)

đặt \(\sqrt{\frac{x+1}{x^2-x+1}}=a;a\ge0\)

tc pt \(a-\frac{2}{a}+1=0\)

\(a\left(1-\frac{1}{a^2}\right)-\frac{1}{a}+1=0\)

\(a\left(1-\frac{1}{a}\right)\left(1+\frac{1}{a}\right)+1-\frac{1}{a}=0\)

\(\left(1-\frac{1}{a}\right)\left(a+2\right)=0\)

\(\Rightarrow a=1\)(a+2>0)

\(\Rightarrow\sqrt{\frac{x+1}{x^2-x+1}}=1\)

\(\Rightarrow x+1=x^2-x+1\)

\(\Rightarrow x^2-2x=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=0\left(tm\right)\end{cases}}\)