a, \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)+  \(\sqrt{\left(\sqrt{3}\right)^2-2\cdot\left(\sqrt{3}\right)\cdot\left(\sqrt{2}\right)+\left(\sqrt{2}\right)^2}\)

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

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

a) \(\sqrt{5+2\sqrt{6}+}\sqrt{5-2\sqrt{6}}\)

=\(\frac{\sqrt{10+4\sqrt{6}}}{\sqrt{2}}+\frac{\sqrt{10-4\sqrt{6}}}{\sqrt{2}}\)

=\(\frac{\sqrt{\left(\sqrt{6+2}\right)}^2}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{6-2}\right)}^2}{\sqrt{2}}\)

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

=\(\frac{\sqrt{2\left(\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(\sqrt{3}-\sqrt{2}\right)}}{\sqrt{2}}\)

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

=\(2\sqrt{3}\)

b )\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

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

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

=\(\frac{-2}{\sqrt{2}}\)

=\(-\sqrt{2}\)

a )\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)

=\(\sqrt{2+3+1+2\sqrt{2.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}}\)

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

=\(\sqrt{2}+\sqrt{3}+1\)

sai đề nhé ở đây, min nó là 16 mà 6 căn 6=14 thôi, mà cái điểm rơi cũng ngộ nữa :))

Nếu bạn đã nói sai thì cho mình giải thử nhé!

Áp dụng BĐT Bunhiacopxky - Cauchy - Schwarz, ta có: 

\(\left(ax+by+cz\right)^2\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)\(\Rightarrow\sqrt{a^2+b^2+c^2}\cdot\sqrt{x^2+y^2+z^2}\ge ax+by+cz\)(với a, b, c, x, y, z là những số dương)

\(\Rightarrow\sqrt{2+18+4}\cdot\sqrt{\frac{8}{a^2}+\frac{9b^2}{2}+\frac{c^2a^2}{4}}\ge\sqrt{2}\cdot\frac{2\sqrt{2}}{a}+3\sqrt{2}\cdot\frac{3b}{\sqrt{2}}+2\cdot\frac{ca}{2}\)

\(\Leftrightarrow\sqrt{24}\cdot\sqrt{\frac{8}{a^2}+\frac{9b^2}{2}+\frac{c^2a^2}{4}}\ge\frac{4}{a}+9b+ca\)(1)

Tương tự ta có: \(\sqrt{24}.\sqrt{\frac{8}{b^2}+\frac{9c^2}{2}+\frac{a^2b^2}{4}}\ge\frac{4}{b}+9c+ab\)(2)

                           \(\sqrt{24}\cdot\sqrt{\frac{8}{c^2}+\frac{9a^2}{2}+\frac{b^2c^2}{4}}\ge\frac{4}{c}+9a+bc\)(3)

Cộng vế theo vế (1), (2) và (3) ta được: \(\sqrt{24}\cdot\left(VT\right)\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}+9\left(a+b+c\right)+ab+bc+ca\)

\(=\left(\frac{4}{a}+a\right)+\left(\frac{4}{b}+b\right)+\left(\frac{4}{c}+c\right)+\left(2a+bc\right)+\left(2b+ca\right)+\left(2c+ab\right)\)\(+6\left(a+b+c\right)\)\(\ge2\sqrt{\frac{4}{a}\cdot a}+2\sqrt{\frac{4}{b}\cdot b}+2\sqrt{\frac{4}{c}\cdot c}+2\sqrt{2abc}+2\sqrt{2abc}+2\sqrt{2abc}\)\(+6\left(a+b+c\right)\)\(=12+6\left(a+b+c+\sqrt{2abc}\right)\ge12+6\cdot10=72\)

\(\Rightarrow VT\ge\frac{72}{\sqrt{24}}=6\sqrt{6}\)

Dấu ''='' xảy ra khi: \(\hept{\begin{cases}a+b+c+\sqrt{2abc}=10\\VT=6\sqrt{6}\end{cases}\Leftrightarrow a=b=c=2}\)

Vậy ta được ĐPCM

\(x+y=4xy\Rightarrow\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}=4\)

\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow4>=\frac{4}{x+y}\Rightarrow x+y>=1\)(bđt svacxo)

\(x^2+y^2>=\frac{\left(x+y\right)^2}{2};xy< =\frac{\left(x+y\right)^2}{4}\)

\(\Rightarrow P=x^2+y^2-xy>=\frac{\left(x+y\right)^2}{2}-\frac{\left(x+y\right)^2}{4}=\frac{\left(x+y\right)^2}{4}>=\frac{1^2}{4}=\frac{1}{4}\)

dấu = xảy ra khi \(x+y=1;x=y\Rightarrow x=y=\frac{1}{2}\left(tm\right)\)

vậy min P là \(\frac{1}{4}\)khi x=y=\(\frac{1}{2}\)

| mặt cừ| :v
Ta có xy-x^2 = 5 
Mà 20-y^2 = 4.5-y^2 = 4(xy-x^)-y^2 = ( - (4x^2 - 4xy +y^2) = - (2x-y)^2 
<=> \sqrt(x+y-5) + (2x-y)^2=0
<=> x+y-5=0 và 2x-y=0 phải xảy ra đồng thời ^^  <=> x=5/3 ; y= 10/3 thé vào pt => Vô nghiệm ^^

x²-yx +5 =0\(\Rightarrow A=y^2-20>=0\)

\(\sqrt{x+y-5}=20-y^2>=0.\)

\(\Rightarrow y^2=20\)

\(\Rightarrow y=2\sqrt{5}\)

và x =y/2 =\(\sqrt{5}\)

và x +y -5 =0 .(vô lí ) .

\(\Rightarrow\)HPT Vô nghiệm .