Lời giải :

Đặt \(\hept{\begin{cases}a+2b+c=x\\a+b+2c=y\\a+b+3c=z\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=-x+5y-3z\\b=x-2y+z\\c=z-y\end{cases}}\)

Thay vào P ta được :

\(P=\frac{-x+5y-3z+3z-3y}{x}+\frac{4x-8y+4z}{y}+\frac{-8z+8y}{z}\)

\(P=-1+\frac{2y}{x}+\frac{4x}{y}-8+\frac{4z}{y}-8+\frac{8y}{z}\)

\(P=-17+\left(\frac{2y}{x}+\frac{4x}{y}\right)+\left(\frac{4z}{y}+\frac{8y}{z}\right)\)

Áp dụng BĐT Cô-si :

\(P\ge-17+2\sqrt{\frac{2y\cdot4x}{x\cdot y}}+2\sqrt{\frac{4z\cdot8y}{x\cdot z}}\)

\(=-17+2\sqrt{8}+2\sqrt{32}\)

\(=-17+12\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{2y}{x}=\frac{4x}{y}\\\frac{4z}{y}=\frac{8y}{z}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x^2=y^2\\z^2=2y^2\end{cases}}\)

Thay a,b,c vào tìm ra dấu "=" nhé. Khá dài và phức tạp đấy.

Ai ti-ck sai ra đây nói chuyện nào ?

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 :

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

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

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

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

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

\(=4-1=3\)

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

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

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

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