Áp dụng bđt Svacxo có

\(\frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\ge\frac{\left(a+b+c\right)^2}{b+c-a+c+a-b+a+b-c}=a+b+c\)

Dấu "=" tại a =b = c

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

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)

\(b,\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{z}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{z}^2}\)

\(=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{x+2\sqrt{xy}+y-z}\)

M đạt max khi \(\frac{1}{M}\) đạt min

\(\frac{1}{M}=1-\frac{1}{x}+\frac{1}{x^2}=\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}+\frac{3}{4}\)

\(\frac{1}{M}\ge\frac{3}{4}\Rightarrow M\le\frac{4}{3}\)

dấu = xảy ra khi x=1/2

bạn trả lời rõ hơn dc ko bạn

xem đúng đề k

ĐKXĐ: \(\hept{\begin{cases}x^2+5x+3\ge0\\x^2+3x+2\ge0\end{cases}}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2+5x+3}=a\\\sqrt{x^2+3x+2}=b\end{cases}\left(a;b\ge0\right)}\)

\(\Rightarrow a^2-b^2=x^2+5x+3-x^2-3x-2=2x+1\)

Pt trở thành

\(a-b=a^2-b^2\)

\(\Leftrightarrow a-b-\left(a-b\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(1-a-b\right)=0\)

\(pt\Leftrightarrow2\sqrt{\frac{x-1}{4}}-6=2\sqrt{\frac{4\left(x-1\right)}{9}}-\frac{1}{3}\)

\(\Leftrightarrow\sqrt{x-1}-6-\frac{4}{3}\sqrt{x-1}+\frac{1}{3}=0\)

\(\Leftrightarrow-\frac{1}{3}\sqrt{x-1}=\frac{17}{3}\)(vô lý )

Vậy pt vô nghiệm

\(=\sqrt{\left(2\sqrt{5}-7\right)^2}-\sqrt{\left(5-2\sqrt{5}\right)^2}\)

\(=7-2\sqrt{5}-5+2\sqrt{5}\)

\(=2\)

giúp mình với