Question

[Đan Cao Thị Linh]

\(\sqrt{x^2+15}=3x-2+\sqrt{x^2+8}\)

\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

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

\(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)

\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)

Answer 1

[kudo shinichi]

\(\frac{\left(\sqrt{x^2+15}-4\right).\left(\sqrt{x^2+15}+4\right)}{\sqrt{x^2+15}+4}=3x-3+\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\sqrt{x^2+8}+3}\)

\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+15}+4}=3\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+8}+3}\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}\right)=0\)

\(\Leftrightarrow3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}=0\)hoặc x=1

Ta có: \(\sqrt{x^2+15}-\sqrt{x^2+8}=3x-2\)

Thấy: VT>0 => VP>0 => x>2/3

Xét \(3+\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+15}+4}=0\)(1)

Ta thấy: với x>2/3 thì VT luôn dương => (1) vô lý

Vậy S={1}

Answer 2

[kudo shinichi]

\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

\(\Leftrightarrow x^2-8=\left(x+3\right)\frac{\left(\sqrt{x^2+1}-3\right)\left(\sqrt{x^2+1}+3\right)}{\sqrt{x^2+1}+3}\)

\(\Leftrightarrow x^2-8=\left(x+3\right)\frac{x^2-8}{\sqrt{x^2+1}+3}\)

\(\Leftrightarrow\left(x^2-8\right)\left(1-\frac{x+3}{\sqrt{x^2+1}+3}\right)=0\)

\(\Leftrightarrow\left(x^2-8\right)\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+3}=0\)

Có \(\sqrt{x^2+1}-x>0\)

\(\Leftrightarrow\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+3}>0\)

\(\Rightarrow x=\pm2\sqrt{2}\)

Vậy...