Ta có (x + |x| + 2016)(y + |y| + 2016) > 2016 với mọi x, y nên không thể tính được P

x+y =0

=> P = 1

x+y=0

=>P=1

\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)

\(\Rightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)

\(\Rightarrow\frac{x}{2015}-\frac{x}{2017}=0\)

\(\Rightarrow x.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)

Mà ta thấy \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow x=0\)

Vậy \(x=0\)

\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)

\(\Leftrightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)

\(\Leftrightarrow x\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)

\(\Leftrightarrow x=0\).Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\ne0\)

Vậy giá trị của x là x=0

\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)

<=>\(\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)=0\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)

<=>\(\left|x\right|.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)

Vì \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow\left|x\right|=0\Rightarrow x=0\)

Vậy x=0

\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)

\(\Rightarrow\left|x.\left(\frac{1}{2015}+\frac{1}{2016}\right)\right|=\left|x.\left(\frac{1}{2016}+\frac{1}{2017}\right)\right|\)

\(\Rightarrow\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)

\(\Rightarrow\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)

Mà \(\frac{1}{2015}+\frac{1}{2016}>\frac{1}{2016}+\frac{1}{2017}\)

=> |x| = 0

=> x = 0

Vậy x = 0

kho....................wa..................troi.......................thi.....................ret.................lanh................wa..................tich............................ung.........................ho..............minh......................cho....................do....................lanh

Ta có: 5x²+5y²+8xy-2x+2y+2=0

=> 4x²+8xy+4y²+x²-2x+1+y²+2y+1=0

=> (2x+2y)²+(x-1)²+(y+1)²=0

=> {2x+2y=0 => x=-y

      {x-1 = 0 => x=1

      {y+1 =0 => y=-1

=> x=1, y=-1

Thay vào biểu thức M, ta có:

M=(1+-1)²⁰¹⁵+(1-2)²⁰¹⁶+(-1+1)²⁰¹⁷=0+1+0=1 (đpcm)

phân tích đẳng thức trên

bài này dễ ợt. Tính đc M=1

giúp mình đi nha mình cần rất rất rất rất ...... gấp đấy !!!!=.='

x = -2014

X = 2014 

ĐUNG

| x + 2015 | + 2016 = 2017 

| x + 2015 | = 2017 - 2016 = 1 

x + 2015 = 1 hoặc x + 2015 = -1 

x = 1 - 2015 hoặc x = -1 - 2015 

x = -2014 hoặc x = - 2016