Lời giải :

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

ĐKXĐ: \(\hept{\begin{cases}2-2x\ge0\\x-1\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge1\end{cases}}\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

Lời giải :

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

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

\(=\left|x+1\right|+\left|x-1\right|\)

\(=\left|x+1\right|+\left|1-x\right|\ge\left|x+1+1-x\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)\left(1-x\right)\ge0\Leftrightarrow-1\le x\le1\)

hình như là \(x\in R\)

\(a,\frac{a+b}{2}\ge\sqrt{ab}\Rightarrow a+b\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\)( luôn đúng ) 

\(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\)

Lời giải :

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{16}-\sqrt{15}}{\left(\sqrt{15}+\sqrt{16}\right)\left(\sqrt{16}-\sqrt{15}\right)}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{16}-\sqrt{15}}{16-15}\)

\(=\frac{\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{16}-\sqrt{15}}{1}\)

\(=\sqrt{16}-\sqrt{1}\)

\(=4-1=3\)

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)

\(=-\left(\sqrt{1}-\sqrt{2}\right)-\left(\sqrt{2}-\sqrt{3}\right)-...-\left(\sqrt{15}-\sqrt{16}\right)\)

\(=-\left(\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{15}-\sqrt{16}\right)\)

\(=-\left(\sqrt{1}-\sqrt{16}\right)=-\left(1-4\right)=3\)