A=\(\frac{u-v}{\sqrt{u}+\sqrt{v}}-\frac{\sqrt{u^3}+\sqrt{v^3}}{u-v}=\frac{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}{\sqrt{u}+\sqrt{v}}-\frac{\left(\sqrt{u}+\sqrt{v}\right)\left(u-\sqrt{u}\sqrt{v}+v\right)}{\left(\sqrt{u}+\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\frac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}=\frac{u-2\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}=\frac{-\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)

2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(-\sqrt{x+2}=-9\)

\(x+2=81\)

\(\Rightarrow x=79\)

3) \(\sqrt{x^2-4x+4}=7x-1\)

\(\sqrt{x^2-2.x.2+2^2}=7x-1\)

\(\sqrt{\left(x-2\right)^2}=7x-1\)

\(x-2=7x-1\)

\(-2=7x-1-x\)

\(-2+1=7x-x\)

\(-1=6x\)

\(-\frac{1}{6}=x\)

\(\Rightarrow x=-\frac{1}{6}\)

\(B=\left|5+3\sqrt{2}\right|+\left|\sqrt{11}-3\sqrt{2}\right|+\frac{11}{\sqrt{11}}\)

\(=5+3\sqrt{2} +3\sqrt{2}-\sqrt{11}+\sqrt{11}\)

\(=5+6\sqrt{2}\)

Lên trên mạng nhá

thế onl sinh ra để vứt đi à

Giải:

CD vuông AB tại H

=> OA vuông CD tại H

=> CD = 2. CH

Tam giác ACB vuông tại C ( vì AB là đường kính)

=> CB^2 =AB^2 - AC^2= 5^2 - 3^2 =16

=> CB = 4

\(\Rightarrow\frac{1}{CH^2}=\frac{1}{AC^2}+\frac{1}{AB^2}=\frac{1}{3^2}+\frac{1}{4^2}=\frac{25}{144}\)

=> \(CH=\frac{12}{5}\Rightarrow CD=2CH=\frac{24}{5}\)

\(\sqrt{x}\)

Ta có: 

\(\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2=2\left(a+b+c\right)\)

=> \(\left(a+b+c\right)^2-6\left(a+b+c\right)\le0\)

=> \(0\le a+b+c\le6.\)

\(T=\frac{a}{a+1}+\frac{b}{b+a}+\frac{c}{c+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}\)

\(=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\le3-\frac{\left(1+1+1\right)^2}{a+b+c+3}\le3-\frac{3^2}{6+3}=2\)

"=" xảy ra <=> \(a=b=c\)và \(a+b+c=6\)<=> \(a=b=c=2\)

Vậy max T = 2 khi và chỉ khi a=b=c =2

Ta co:

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(z+\frac{1}{z}\right)^2\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{3}\ge\frac{\left(1+\frac{9}{x+y+z}\right)^2}{3}=\frac{100}{3}\)

Dau '=' xay ra khi \(x=y=z=\frac{1}{3}\)

Vay \(A_{min}=\frac{100}{3}\)khi \(x=y=z=\frac{1}{3}\)

\(DK:\hept{\begin{cases}x\ge2\\y\ge3\\z\ge5\end{cases}}\)

\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\)