Aps dụng tính chất dãy tỉ số bằng nhau Ta có:

\(\frac{1+2+3}{x-1+y-2+z-3}=\frac{1+2+3}{x+y+z-1-2-3}=\frac{1+4+9}{x+2y+3z-\left(-4\right)}=\frac{ }{ }\)

=\(\frac{14}{56+4}=\frac{14}{60}=\frac{7}{30}\)

\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{30}\)\(\Rightarrow\)x-1=\(\frac{30}{7}\)\(\Rightarrow\)x=\(\frac{37}{7}\)

\(\Rightarrow\)\(\frac{2}{y-2}=\frac{7}{30}\Rightarrow y-2=\frac{60}{7}\)\(\Rightarrow\)y=\(\frac{74}{7}\)

\(\Rightarrow\)\(\frac{3}{z-3}=\frac{7}{30}\Rightarrow z-3=\frac{90}{7}\)\(\Rightarrow\)x=\(\frac{111}{7}\)

Aps dụng tính chất dãy tỉ số bàng nhau, ta có:

\(\frac{1+2+3}{x-1+2y-2+z-3}=\frac{1+4+9}{x-1+2y-4+3z-9}\)=\(\frac{14}{x+2y+3z-1-2-3}=\frac{14}{56-1-2-3}=\frac{14}{50}=\frac{7}{25}\)

\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{25}\Rightarrow x=\frac{32}{7}\)

\(\Rightarrow\)\(\frac{4}{2y-4}=\frac{7}{25}\Rightarrow y=\frac{64}{7}\)

\(\Rightarrow\)\(\frac{9}{3z-9}=\frac{7}{25}\Rightarrow z=\frac{96}{7}\)

Ta có:  \(\frac{4}{3x-2y}=\frac{3}{2z-4x}=\frac{2}{4y-3z}\)

\(\Rightarrow\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)

\(=\frac{4.\left(3x-2y\right)}{4.4}=\frac{3.\left(2z-4x\right)}{3.3}=\frac{2.\left(4y-3z\right)}{2.2}\)

\(=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\frac{0}{29}=0\)

\(\Rightarrow\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}\)\(\Rightarrow\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}\)\(\Rightarrow12x=8y=6z\)

= \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=\frac{x+y-z}{\frac{1}{12}+\frac{1}{8}-\frac{1}{6}}=\frac{-10}{\frac{1}{24}}=-10.24=-240\)

\(\Rightarrow\begin{cases}x=-240.\frac{1}{12}=-20\\y=-240.\frac{1}{8}=-30\\z=-240.\frac{1}{6}=-40\end{cases}\)

Vậy x = -20; y = -30; z = -40

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

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

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

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

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

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

=>A=\(\frac{x^2-y^2+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

       \(=\left[\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}:\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

a)

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)

Áp dụng tc của dãy tỉ só bằng nhau

\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)

=> x=2.10=20

    y=5.10=50

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)

     \(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)

Mà 2;5 cùng dấu

=> x; y cùng dấu

Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)

a) Ta có: \(\frac{x}{2}\) = \(\frac{y}{5}\) và 3x-y = 10

=> \(\frac{3x}{6}\) = \(\frac{y}{5}\) và 3x-y = 10

Áp dụng tính chất của dãy tỉ số bằng nhau ta có: 

\(\frac{3x}{6}\) = \(\frac{y}{5}\) = \(\frac{3x-y}{6-5}\) = \(\frac{11}{1}\) = 11

=> x= \(\frac{11.6}{3}\) = 22

=> y= 11.5= 55

Vậy x= 22

       y= 55