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 ...

Mình chỉ hướng dẫn giải thôi nhá chứ nhiều bài quá

a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow x=5k;y=7k\)

Thay x.y=315 => 5k.7k=315 <=> 35k²=315 => k²=9 => k=3

x=5.3=15 ; y=7.3=21

b) 5x=9y<=> \(\frac{x}{9}=\frac{y}{5}\)

Theo TCDTSBN ta có : \(\frac{x}{9}=\frac{y}{5}=\frac{2x+3y}{2.9+3.5}=\frac{-33}{33}=-1\)

x/9=-1=>x=-9 ; y/5=-1=>y=-5

các bài còn lại tương tự b 

1, \(4x=5y\) 

mà  \(y-2x=-5\)

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

\(\Rightarrow\left(\frac{y+5}{2}\right).4=5y\)

\(\Rightarrow\frac{4y+20}{2}=5y\)

\(\Rightarrow2y+10=5y\)

\(\Rightarrow10=3y\)

\(\Rightarrow y=\frac{10}{3}\)

\(\Rightarrow x=\frac{y+5}{2}=\frac{\frac{10}{3}+5}{2}=\frac{\frac{25}{3}}{2}=\frac{25}{6}\)

Vậy \(x=\frac{25}{6};y=\frac{10}{3}\)

b, \(\frac{x}{3}=\frac{y}{4}\)

mà \(xy=192\)

Gọi \(x=3k\)

        \(y=4k\)

\(\Rightarrow3k.4k=192\)

\(\Rightarrow12.k^2=192\)

\(\Rightarrow k^2=\frac{192}{12}\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k^2=4^2\)

\(\Rightarrow k=4\)

\(\Rightarrow x=3k=3.4=12\)

\(\Rightarrow y=4k=4.4=16\)

Vậy \(x=12;y=16\)

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{2x+y}{2\left(x+y\right)}-\dfrac{x+2y}{x-y}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{2x^2-2xy+xy-y^2}{2\left(x+y\right)\left(x-y\right)}-\dfrac{2\left(x+2y\right)\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{2x^2-xy-y^2-2\left(x^2+xy-2y^2\right)}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)

\(=\dfrac{2x^2-xy-y^2-2x^2-2xy+4y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)

\(=\dfrac{-3xy+3y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)

\(=\dfrac{-9xy+9y^2-8x}{6\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)

\(=\dfrac{-9x^2y+9xy^2-8x^2+30\left(x^2-y^2\right)}{6x\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{-9x^2y+9xy^2+22x^2-30y^2}{6x\cdot\left(x-y\right)\left(x+y\right)}\)