\(P=2-\sqrt{x^2-x}\)

để P max thì \(2-\sqrt{x^2-x}\)max hay \(\sqrt{x^2-x}\)min 

Mà \(\sqrt{x^2-x}\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

nên P max = 2 \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(\left(5\sqrt{3}+3\sqrt{5}\right):15\)

\(=\sqrt{5}.\sqrt{3}\left(\sqrt{5}+\sqrt{3}\right):15\)

\(=\sqrt{15}\left(\sqrt{5}+\sqrt{3}\right):15=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{15}}\)

a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)

\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)

b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)

\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)

\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)

\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)

c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=27+12\sqrt{5}+12\sqrt{5}\)

\(=27+24\sqrt{5}\)

d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)

e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)

\(=4+2\sqrt{3}-2\sqrt{3}+4\)

= 8

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)

\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

= 14

a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)

= 9 (đpcm)

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

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

\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)

Tìm GTNN của P=a^7+b^7+c^7 biết a^3b^3+b^3c^3+c^3a^3>=1 - Sasu ka

Làm mẫu 1 phần :

a) \(|3x-1|+|x-1|=4\left(1\right)\)

Ta có: \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

             \(x-1=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

3x-1 x-1 1/3 1 0 0 - - - + + + +

+) Với \(x< \frac{1}{3}\Rightarrow\hept{\begin{cases}3x-1< 0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=1-3x\\|x-1|=1-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(1-3x\right)+\left(1-x\right)=4\)

\(2-4x=4\)

\(4x=-2\)

\(x=\frac{-1}{2}\)( chọn )

+) Với \(\frac{1}{3}\le x< 1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=1-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :
\(\left(3x-1\right)+\left(1-x\right)=4\)

\(2x=4\)

\(x=2\)( chọn )

+) Với \(x\ge1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1>0\end{cases}\Rightarrow}\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=x-1\end{cases}\left(4\right)}\)

Thay (4) vào (1) ta được :

\(\left(3x-1\right)+\left(x-1\right)=4\)

\(4x-2=4\)

\(4x=6\)

\(x=\frac{3}{2}\)( chọn )

Vậy \(x\in\left\{\frac{-1}{2};2;\frac{3}{2}\right\}\)