\(A=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=x-\sqrt{xy}+y\)

\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)

\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)

\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)

a: \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{4xy}{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

b: \(=\sqrt{x}+\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)-2\sqrt{y}\)

\(=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0\)

c: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

`[\sqrt{27}-\sqrt{15}]/[3-\sqrt{5}]+4/[2+\sqrt{3}]-6/\sqrt{3}`

`=[\sqrt{3}(3-\sqrt{5})]/[3-\sqrt{5}]+[4(2-\sqrt{3})]/[4-3]-[2\sqrt{3}.\sqrt{3}]/\sqrt{3}`

`=\sqrt{3}+8-4\sqrt{3}-2\sqrt{3}`

`=8-5\sqrt{3}`

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`[x-y]/[\sqrt{x}+\sqrt{y}]-[x\sqrt{y}+y\sqrt{x}]/\sqrt{xy}`    `ĐK:  x,y > 0`

`=[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]/[\sqrt{x}+\sqrt{y}]-[\sqrt{xy}(\sqrt{x}+\sqrt{y})]/\sqrt{xy}`

`=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}`

`=-2\sqrt{y}`

a: \(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)x+\left(\sqrt{3}-1\right)y=\sqrt{3}\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)^2\cdot x+\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)y=\sqrt{3}\left(\sqrt{3}+1\right)\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}\right)+2y=3+\sqrt{3}\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}+2\sqrt{3}\right)=3+\sqrt{3}+3\sqrt{3}+1\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\2y=2\sqrt{3}-3\sqrt{3}-1=-\sqrt{3}-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{-\sqrt{3}-1}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}x\sqrt{3}+y\sqrt{2}=1\\x\sqrt{2}+y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\sqrt{6}+2y=\sqrt{2}\\x\sqrt{6}+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y-3y=\sqrt{2}-3\\x\sqrt{3}+y\sqrt{2}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=\sqrt{2}-3\\x\sqrt{3}=1-y\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-\sqrt{2}\left(3-\sqrt{2}\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-3\sqrt{2}+2=3-3\sqrt{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x=\sqrt{3}-\sqrt{6}\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=\left(x+1\right)\left(y-3\right)\\\left(x-5\right)\left(y+4\right)=\left(x-4\right)\left(y+1\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-2y-y+2=xy-3x+y-3\\xy+4x-5y-20=xy+x-4y-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2x-y+2=-3x+y-3\\4x-5y-20=x-4y-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2x-y+3x-y=-3-2=-5\\4x-5y-x+4y=-4+20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2y=-5\\3x-y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-15\\3x-y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-5y=-15-16=-31\\x-2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{31}{5}\\x=-5+2y=-5+\dfrac{62}{5}=\dfrac{37}{5}\end{matrix}\right.\)

bài này dễ nhưng bạn phải chứng minh bđt này đã:

\(\frac{1}{a+b+c+d}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\)

với a;b;c;d là các số dương

bạn có thể cm bđt trên bằng cách biến đổi tương đương hoặc cm bđt Schwat (Sơ-vác)

Mình là 1 phần tử đại diện còn lại là hoàn toàn tt nhé 

ta có \(\frac{1}{3\sqrt{x}+3\sqrt{y}+2\sqrt{z}}=\frac{1}{2\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{y}+\sqrt{z}\right)+\left(\sqrt{x}+\sqrt{z}\right)}\)

\(\le\frac{1}{16}\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{x}+\sqrt{z}}\right)\)

Tương tự ta cm được 

\(VT\le\frac{1}{16}.4\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{z}+\sqrt{x}}\right)\)\(=\frac{1}{4}.3=\frac{3}{4}\)

dấu "=" khi x=y=z

Bài 2: 

a: Ta có: \(\sqrt{\sqrt{5}-x\sqrt{3}}=\sqrt{8+2\sqrt{15}}\)

\(\Leftrightarrow\sqrt{5}-x\sqrt{3}=8+2\sqrt{15}\)

\(\Leftrightarrow x\sqrt{3}=\sqrt{5}-8-2\sqrt{15}\)

\(\Leftrightarrow x=\dfrac{\sqrt{15}-8\sqrt{3}-6\sqrt{5}}{3}\)

b: Ta có: \(\sqrt{2+\sqrt{\sqrt{x}+3}}=3\)

\(\Leftrightarrow\sqrt{\sqrt{x}+3}=7\)

\(\Leftrightarrow\sqrt{x}=46\)

hay x=2116

Kiểm tra lại đề bài đi em, chỗ CMR đó

Đặt \(\sqrt[3]{x^2}=m\Leftrightarrow x^2=m^3;\sqrt[3]{y^2}=n\Leftrightarrow y^2=n^3\)

Thay vào biểu thức:

\(\Leftrightarrow\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\\ \Leftrightarrow m^3+n^3+mn\left(m+n\right)+2\sqrt{\left(m^3+m^2n\right)\left(n^3+mn^2\right)}=a^2\\ \Leftrightarrow m^3+n^3+mn\left(m+n\right)+2\sqrt{m^2n^2\left(m+n\right)}=a^2\\ \Leftrightarrow m^3+n^3+3mn\left(m+n\right)=a^2\\ \Leftrightarrow\left(m+n\right)^3=a^2\\ \Leftrightarrow m+n=\sqrt[3]{a^2}\\ \Leftrightarrow\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)

Em chắc chắn là đề bài đúng chứ? Trước khi nhìn kĩ lại?