Có em nhé.

Ta có:

\(P=\dfrac{5}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{5}{x^2+y^2}+\dfrac{5}{2xy}+\dfrac{1}{2xy}\\ =5\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)

Với hai số dương \(x;y\) , bằng cách khai triển tương đương hai vế ta dễ dàng chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (BĐT Cauchy-Schwarz)

Áp dụng vào biểu thức P ta có:

\(P\ge5\left(\dfrac{4}{x^2+2xy+y^2}\right)+\dfrac{1}{2xy}\\ \ge5\left(\dfrac{4}{\left(x+y\right)^2}\right)+\dfrac{2}{\left(x+y\right)^2"cosy"}\\ \ge\dfrac{5.4}{3^2}+\dfrac{2}{3^2}=\dfrac{22}{9}\)

Dấu \('='\) xảy ra khi \(x=y=\dfrac{3}{2}\)

\(giúp\) \(mình\) \(với\) ☹

☹☹☹☹

Ta có:

\(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)\ge4\)

\(\sqrt{ab}+\sqrt{a}+\sqrt{b}-3\ge0\)

Áp dụng BĐT Cô si, ta có:

\(\sqrt{ab}\le\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)

\(\Rightarrow0\le\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}+\left(\sqrt{a}+\sqrt{b}\right)-3\)

\(\Rightarrow0\le\left(\sqrt{a}+\sqrt{b}\right)^2+4\left(\sqrt{a}+\sqrt{b}\right)-12\)

\(\Rightarrow0\le\left(\sqrt{a}+\sqrt{b}-2\right)\left(\sqrt{a}+\sqrt{b}+6\right)\)

Vì \(\sqrt{a}+\sqrt{b}+6>0\)

\(\Rightarrow\sqrt{a}+\sqrt{b}-2\ge0\Rightarrow\sqrt{a}+\sqrt{b}\ge2\)

Áp dụng BĐT Svácxơ, ta có:

\(P=\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge\dfrac{\left(a+b\right)^{^2}}{a+b}=a+b\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2}=\dfrac{2^2}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)

a)

Đk: \(x+5\ge0\Rightarrow x\ge-5\)

\(\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}=12\\ \Leftrightarrow\sqrt{4\left(x+5\right)}-2\sqrt{x+5}+\sqrt{9\left(x+5\right)}=12\\ \Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=12\\ \Leftrightarrow3\sqrt{x+5}=12\\ \Leftrightarrow\sqrt{x+5}=4\\ \Leftrightarrow x+5=16\\ \Rightarrow x=11\)

b.

\(\sqrt{x^2-10x+25}=6\)

Đk: \(x^2-10x+25=\left(x-5\right)^2\ge0;\forall x\inℝ\)

\(\sqrt{x^2-10x+25}=6\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=6\\ \Rightarrow|x-5|=6\)

\(\Rightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)

Đs....