1) Áp dụng tích chất dãy tỉ số bằng nhau ta có:

\(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y-x+y}{2015-2017}=\frac{2y}{-2}\)

\(=-y\)

\(\Rightarrow xy=-2016y;x+y=-2015y;\)

\(x-y=-2017y\)

\(\Rightarrow-2016y-xy=0\)

\(\Rightarrow y\left(-2016-x\right)=0\)

\(\Rightarrow\orbr{\orbr{\begin{cases}y=0\\-2016-x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}y=0\\x=-2016\end{cases}}\)

\(+) \)\(y=0\Rightarrow0+x=-2015.0=0\Rightarrow x=0\)

\(+) \)\(x=-2016\Rightarrow-2016-y=-2017y\Rightarrow-2016\)

Vậy +) x=y=0

       +) x=-2016;y=1

2) Có: \(\frac{2x+2}{3}=\frac{x+1}{1,5};\frac{4z+2}{5}=\frac{z+0,5}{1,25};\frac{3y-1}{4}=\frac{y-\frac{1}{3}}{\frac{4}{3}}\)

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

\(\frac{x+1}{1,5}=\frac{y-\frac{1}{3}}{\frac{4}{3}}=\frac{z+0,5}{1,25}=\frac{x+y+z+\left(1-\frac{1}{3}+0,5\right)}{1,5+\frac{4}{3}+1,25}=\frac{7+\frac{7}{6}}{\frac{49}{12}}=2\)

Suy ra: \(x+1=2.1,5=3\Rightarrow x=2\)

             \(y-\frac{1}{3}=2.\frac{4}{3}=\frac{8}{3}\Rightarrow y=3\)

            \(z+0,5=2.1,25=2,5\Rightarrow z=2\)

Vậy x=2;y=3;z=2.

Câu 1 :

Áp dụng t/c dãy TSBN ta có : \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y+x-y}{2015+2017}=\frac{x}{2016}\)

\(\Rightarrow\frac{xy}{2016}=\frac{x}{2016}\)=> xy=x => xy-x=0 => x(y-1)=0 => x=0 hoặc y=1

+) Nếu x=0 => \(\frac{0+y}{2015}=\frac{0.y}{2016}\Rightarrow\frac{y}{2015}=0\Rightarrow y=0\)

+) Nếu y=1 => \(\frac{x+1}{2015}=\frac{x.1}{2016}\)=> 2016(x+1)=2015x => 2016x+2016 = 2015x => x=-2016

             Vậy ...

Câu 2 :

Áp dụng t/c dãy TSBN ta có : \(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4z+2}{5}=\frac{6.\left(2x+2\right)+4.\left(3y-1\right)+3.\left(4z+2\right)}{3.6+4.4+5.3}\)

                                             \(=\frac{12\left(x+y+z\right)+14}{49}=\frac{12.7+14}{49}=2\)

Từ  \(\frac{2x+2}{3}=2\Rightarrow2x+2\Rightarrow6\Rightarrow2x=4\Rightarrow x=2\)

Tương tự tìm đc y=3 và z=2

            Vậy ...

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

a) \(\frac{1}{y}+\frac{x}{4}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{y}=\frac{1}{2}-\frac{x}{4}\)

\(\Rightarrow\frac{1}{y}=\frac{2-x}{4}\)

\(\Leftrightarrow\left(2-x\right).y=4\)

Do \(x,y\inℤ\Rightarrow2-x,y\inℤ\)

nên \(2-x,y\) là các cặp ước của 4

Ta có bảng giá trị :

Vậy : \(\left(x,y\right)\in\left\{\left(1,4\right);\left(3,-4\right);\left(0,-2\right);\left(4,2\right);\left(-2,1\right);\left(6,-1\right)\right\}\)

b) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{y}{4}\)

\(\Rightarrow\frac{5}{x}=\frac{1-2y}{8}\)

\(\Leftrightarrow x.\left(1-2y\right)=40\)

Nhận xét x,y và lập bảng giá trị tương tự câu a).

Đặt \(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z=k\)

Áp dụng TC DTSBN ta có :

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

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\z+x+1=2y\\x+y-2=2z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)

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

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)

\(x:z=\frac{2}{3}:\frac{1}{2}=\frac{4}{3}\Rightarrow x=\frac{4}{3}.z\)

\(z:y=1:\frac{4}{7}=\frac{7}{4}\Rightarrow z=y.\frac{7}{4}\)

\(\Rightarrow y+z=y+y.\frac{7}{4}=66\)

\(y.\frac{11}{4}=66\Rightarrow y=24\)

\(\Rightarrow z=24.\frac{7}{4}=42\)

\(\Rightarrow x=42.\frac{4}{3}=56\)

a, \(\frac{x-3}{y-2}=\frac{3}{2}\)và \(x-y=4\)

Theo bài ra ta có : 

\(\frac{x-3}{y-2}=\frac{3}{2}\Leftrightarrow2x-6=3y-6\Leftrightarrow2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)

Áps dụng tính chất dãy tỉ số bằng nhau ta đc :

\(\frac{x}{3}=\frac{y}{2}=\frac{x-y}{3-2}=\frac{4}{1}=4\)

\(\frac{x}{3}=4\Leftrightarrow x=12\)

\(\frac{y}{2}=4\Leftrightarrow y=8\)

Tương tự với b thôi bn.

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)

Suy ra

\(x+y+z=\frac{1}{2}\)(1)

\(y+z+1=2x\)(2)

\(x+z+2=2y\)(3)

\(x+y-3=2z\)(4)

(2)-(1) ta có

\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)

\(y=-z\)

\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)

\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)

\(z=-\frac{5}{6}\)