a, \(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right):\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)ĐK : x >= 0 ; \(x\ne1\)

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

b, \(F=\frac{5}{2}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\)

ĐK : x > 0 , x khác 1

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

Để bthuc = 5/2 thì \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\left(tm\right)\)

HPT<=>\(\hept{\begin{cases}2\left(u+v\right)+v^2+2uv+u^2=15\\u^2+v^2=5\end{cases}}\)

\(< =>\hept{\begin{cases}\left(u+v+1\right)^2=16\\u^2+v^2=5\end{cases}}\)

\(< =>\hept{\begin{cases}u+v=3\\u^2+v^2=5\end{cases}or\hept{\begin{cases}u+v=-5\\u^2+v^2=5\end{cases}}}\)

đến đến thì dễ r haaaa

\(\hept{\begin{cases}uv+u+v=5\\u^2+v^2=5\end{cases}}\)

\(u^2+v^2=\left(u+v\right)^2-2uv=\left(u+v\right)^2-2\left[5-\left(u+v\right)\right]\)

\(=\left(u+v\right)^2+2\left(u+v\right)-10=5\)

\(\Leftrightarrow\left(u+v\right)^2+2\left(u+v\right)-15=0\)

\(\Leftrightarrow\left(u+v+5\right)\left(u+v-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}u+v=-5\\u+v=3\end{cases}}\)

- \(u+v=-5\Rightarrow uv=10\)

\(u,v\)là hai nghiệm của phương trình: \(x^2+5x+10=0\)(1)

mà \(x^2+5x+10=x^2+2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2+\frac{15}{4}=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\)

nên phương trình (1) vô nghiệm. 

- \(u+v=3\Rightarrow uv=2\)

\(u,v\)là hai nghiệm của phương trình \(x^2-3x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy \(\left(u,v\right)\in\left\{\left(1,2\right),\left(2,1\right)\right\}\).

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

\(ĐKXĐ:-1\le x\le1\)

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

\(\left(3\sqrt{1-x}\sqrt{1+x}-\frac{3}{2}\right)+\left(\sqrt{1+x}-x-\frac{1}{2}\right)=0\)

\(\frac{9\left(1-x\right)\left(1+x\right)-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-\left(x+\frac{1}{2}\right)^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)

\(\frac{9-9x^2-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-x^2-x-\frac{1}{4}}{\sqrt{1+x}+x+\frac{1}{2}}=0\)

\(\frac{\frac{27}{4}-9x^2}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)

\(\frac{9\left(\frac{3}{4}-x^2\right)}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)

\(\left(\frac{3}{4}-x^2\right)\left(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}\right)=0\)

\(\orbr{\begin{cases}\frac{3}{4}-x^2=0\\\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}=0\left(KTM\right)\end{cases}< =>x=\frac{\sqrt{3}}{2}\left(TM\right)}\)

\(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}>0\)nên pt ktm

\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{ab}}{b}\)

Bài 1 : a, Theo BĐT Cauchy Schwarz dạng Engel 

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

Dấu ''='' xảy ra khi \(a=b=c\)

b, \(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{bc+ab}+\frac{c^4}{ac+bc}\)(1)

Theo BĐT Cauchy Schwarz dạng Engel \(\left(1\right)\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\)

Theo BĐT phụ \(a^2+b^2+c^2\ge ab+bc+ca\)( bạn nhân 2 vào 2 vế rồi tự cm nhé )

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

Dấu ''='' xảy ra khi a=b=c 

chủ yếu là xài bđt Cauchy-Schwarz dạng Engel nhé 

1a. Áp dụng bđt Cauchy-Schwarz dạng Engel : 

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

=> đpcm . Dấu "=" xảy ra <=> a=b=c > 0

hàm số \(y=\left|x-1\right|+2\)