Ai giúp em với ạ !!!!!!!!

a) Ta có: \(\left|x\right|=\left|y\right|\)

 \(\Rightarrow\left|x\right|=\left[{}\begin{matrix}-x\\x\end{matrix}\right.\)

\(\Rightarrow\left|y\right|=\left[{}\begin{matrix}y\\-y\end{matrix}\right.\)

Mà \(x< 0;y>0\)

\(\Rightarrow x+y=0\)

A∈∅

\(A^2=\left(\frac{y}{z}+\frac{z}{y}\right)^2=\frac{y^2}{z^2}+\frac{z^2}{y^2}+2\cdot\frac{y}{z}\cdot\frac{z}{y}=\frac{y^2}{z^2}+\frac{z^2}{y^2}+2\)

\(B^2 = \left(\frac{x}{z} + \frac{z}{x}\right)^2 = \frac{x^2}{z^2} + \frac{z^2}{x^2} + 2\)

\(C^2 = \left(\frac{x}{y} + \frac{y}{x}\right)^2 = \frac{x^2}{y^2} + \frac{y^2}{x^2} + 2\)

Do đó: \(A^2+B^2+C^2\)

\(=\frac{y^2}{z^2}+\frac{z^2}{y^2}+2+\frac{x^2}{y^2}+\frac{y^2}{x^2}+2+\frac{x^2}{z^2}+\frac{z^2}{x^2}+2\)

\(=\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{x^2}{z^2}+\frac{z^2}{x^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+6\quad\)

\(A \cdot B = \left(\frac{y}{z} + \frac{z}{y}\right)\left(\frac{x}{z} + \frac{z}{x}\right) = \frac{xy}{z^2} + \frac{y}{x} + \frac{x}{y} + \frac{z^2}{xy}\)

=>A*B*C\(=\left(\frac{xy}{z^2}+\frac{z^2}{xy}+\frac{x}{y}+\frac{y}{x}\right)\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(=\frac{x^2}{z^2}+\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{z^2}{x^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\)

\(=\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{x^2}{z^2}+\frac{z^2}{x^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+2\quad\)

\(A^2+B^2+C^2-ABC\)

=6-2

=4

a: \(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

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

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

Ta có: \(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\)

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

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

\(=\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{\left(x+y\right)_{}\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(\) \(=\frac{x^2-\sqrt{xy}\left(x+y\right)-y^2-x^2+y^2}{\sqrt{xy}\left(x-y\right)}=\frac{-\left(x+y\right)}{x-y}\)

b: Thay x=3; \(y=4+2\sqrt3\) vào A, ta được:

\(A=\frac{-\left(3+4+2\sqrt3\right)}{3-\left(4+2\sqrt3\right)}=\frac{-7-2\sqrt3}{-2\sqrt3-1}=\frac{7+2\sqrt3}{2\sqrt3+1}\)

\(=\frac{\left(7+2\sqrt3\right)\left(2\sqrt3-1\right)}{12-1}=\frac{14\sqrt3-7+12-2\sqrt3}{11}=\frac{12\sqrt3+5}{11}\)

\(\frac{x}{y}=\frac23\)

=>\(\frac{x}{2}=\frac{y}{3}=k\)

=>x=2k; y=3k

\(A=\frac{3x+5y}{7x-2y}=\frac{3\cdot2k+5\cdot3k}{7\cdot2k-2\cdot3k}=\frac{6k+15k}{14k-6k}=\frac{21}{8}\)

\(B=\frac{x^2-xy+y^2}{x^2+xy+y^2}\)

\(=\frac{\left(2k\right)^2-2k\cdot3k+\left(3k\right)^2}{\left(2k\right)^2+2k\cdot3k+\left(3k\right)^2}=\frac{4k^2-6k^2+9k^2}{4k^2+6k^2+9k^2}=\frac{7}{19}\)

TH1: \(x+y+z+t\ne0\) 

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)

TH1: \(x+y+z+t=0\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)

Tham khảo: https://hoc24.vn/cau-hoi/cho-bieu-thuc-pdfracxyztdfracyztxdfracztxydfractxyz-tinh-gia-tri-bieu-thuc-p-biet-dfracxyztdfracyzt.3023321885549

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Lời giải:
Nếu $x+y+z+t=0$ thì:

$P=\frac{-(z+t)}{z+t}+\frac{-(t+x)}{t+x}+\frac{-(x+y)}{x+y}+\frac{-(y+z)}{y+z}$

$=-1+(-1)+(-1)+(-1)=-4$

Nếu $x+y+z+t\neq 0$ thì áp dụng TCDTSBN:

$\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3}$

$\Rightarrow 3x=y+z+t; 3y=z+t+x; 3z=t+x+y; 3t=x+y+z$

$\Rightarrow x=y=z=t$

$\Rightarrow P=1+1+1+1=4$