\(\left(\sqrt{45}-\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(\sqrt{9.5}-\sqrt{9.7}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=-3\left(\sqrt{5}-\sqrt{7}\right)^2\)

Ta có:

\(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}\cdot\frac{1+y}{8}\cdot\frac{1+z}{8}}=\frac{3}{4}x\)

Tương tự:

\(\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge\frac{3}{4}y\)

\(\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge\frac{3}{4}z\)

\(\Rightarrow VT+\frac{3}{4}+\frac{1}{4}\left(x+y+z\right)\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Leftrightarrow VT\ge\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{1}{2}\cdot3\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi: x = y = z = 1

Ta có : \(B=\frac{2\sqrt{x}-x}{\sqrt{x}-2}=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-2}\)

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

Mình chưa vẽ hình nhưng mà câu c bạn có sai không? Tại vì bạn ghi thế thì có khác gì chứng minh AK=AD đâu. Bạn xem lại nhá 

\(\frac{2}{AK}=\frac{1}{AD}+\frac{1}{AE}\) nhá

132-79=

ta có :

\(\frac{a^3+b^3}{a^2+ab+b^2}=\frac{2a^3}{a^2+ab+b^2}+\frac{b^3-a^3}{a^2+ab+b^2}=\frac{2a^3}{a^2+ab+b^3}+b-a\)

tương tự rồi cộng theo vế : 

\(LHS\ge2\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)

áp dụng bđt cô si

 \(\frac{a^3}{a^2+ab+b^2}+\frac{a^2+ab+b^2}{9}+\frac{1}{3}\ge\frac{3a}{3}=a\)

tương tự rồi cộng theo vế 

\(2\left(\frac{a^3}{a^2+ab+b^2}+...\right)\ge a+b+c-1-\frac{2\left(a^2+b^2+c^2+ab+bc+ca\right)}{9}\)

\(\ge\frac{2\left(9-a^2-b^2-c^2-ab-bc-ca\right)}{9}\)

đến đây chịu :)))))

đk có nghiệm \(S^2\ge4P\)

a, Ta có : \(x=25\Rightarrow\sqrt{x}=\sqrt{25}=5\)

\(\Rightarrow Q=\frac{5-1}{5+1}=\frac{4}{6}=\frac{2}{3}\)

b, \(P=\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}-\frac{4}{\sqrt{x}}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{4}{\sqrt{x}}\)

\(=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}=\frac{2x-2}{\sqrt{x}}\)

c, Ta có : \(P.Q.\sqrt{x}< 8\)hay \(\frac{2x-2}{\sqrt{x}}.\sqrt{x}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)< 8\)

\(\Leftrightarrow\frac{2\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< 8\Leftrightarrow2\left(\sqrt{x}-1\right)^2< 8\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2< 4\Leftrightarrow\sqrt{x}-1< 2\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)