người ko xem kĩ sẽ khó làm được (VIOLYMPIC)

ta có x=0 (trong VIOLYMPIC)

thử: x/2+x/4+x/2016=x/3+x/5=x/2017

     => 0/2+0/4+0/2016=0/3+0/5=0/2017

     =>   0+0+0 = 0+0 =0             đúng 1000%

ủng hộ nhé

X = 0

Chắc chắn đúng luôn.

Mình vừa thi xong nè 300 điểm đó!

k cho mình nha!

Mai mốt có bài j khó nói với mình , mình chỉ cho.

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

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

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

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

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

\(\frac{x}{2015}+\frac{x}{2016}+\frac{x}{2017}-\frac{x}{2018}\)\(=0\)=> \(x\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}\right)=0\)

Dễ thấy biếu thức trong ngoặc khác 0 nên \(x=0\).

\(x=0\)

Ta có : \(x^4-7x^2+y^2+16=2xy\)

=> \(\left(x^2-8x^2+16\right)+\left(x^2-2xy+y^2\right)=0\)

=> \(\left(x-4\right)^2+\left(x-y\right)^2=0\)

Vì \(\left(x-4\right)^2\ge0 \forall x ,\left(x-y\right)^2 \ge0 \forall x,y \)

=> \(\left(x-4\right)^2+\left(x-y\right)^2\ge0 \forall x,y\)

=> \(\hept{\begin{cases}x-4=0\\x-y=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=y=4\end{cases}}}\)

Thay vào \(A=4^{2016}.4^{2017}-4^{2017}.4^{2016}+4+4=8\)

Vậy A=8

https://olm.vn/thanhvien/nguyentrangth8 bạn giỏi thế

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

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

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

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

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

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

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