Question

[Làm gì mà căng]

\(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-16\)

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

Answer 1

[Nguyễn Thị Mát]

\(ĐKXĐ:x\ge-1\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}\) ta được 

\(a^2-a-20=0\Rightarrow\orbr{\begin{cases}a=5\\a=-4\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)

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

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

\(\Rightarrow x=3\)

Answer 2

[Nguyễn Thị Mát]

b ) \(ĐKXĐ:x\ge0\)

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

Đặt \(\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}\)

\(\Rightarrow a+a^2-2=0\Rightarrow\orbr{\begin{cases}a=1\\a=-2\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{x+1}+\sqrt{x}=1\)

Mà \(x\ge0\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\sqrt{x+1}\ge1\end{cases}\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1}\)

Dấu " = " xảy ra khi và chỉ khi \(x=0\)