=2015 

k mk nha♥

Bằng 2015 nha !!!

x=2015

1,-2015

2,50

3,-2015

thiện xạ 5a3 có thể giải chi tiết ra đc k? Mk cần cách lm

2) 1+2+3+...+x=1275

  Có SSH là: (x+1):1+1=x(SH)

  => (x+1).x:2=1275

=>(x+1).x=1275.2

=>(x+1).x=2550

=>(x+1).x=51.50

=>x=50

3) |x+2015|+|x+2016|=1

Ta thấy  |x+2015| và  |x+2016| > hoặc = 0 với mọi x

=> 1= 0+1=1+0

+) x+2015=0=>x=-2015

     x+2016=1=>x=-2015

+) x+2015=1=>x=-2014

     x+2016=0=> x=-2016

Vậy xE{...}

Tâ có :

B = 2014 x 2016 + 2015

B = 2014 x 2016 + 2016 - 1

B = 2016 x ( 2014 + 1 ) - 1

B = 2016 x 2015 - 1

Vậy A = B

2016 x 2015 - 1 = 2014 x 2016 + 2015.

A=4062239

B=4062239

A=B

Ta có \(|x-y+3|\ge0\forall x,y\)

\(2015\left(2y-3\right)^{2016}\ge0\forall y\)

\(\Rightarrow\hept{\begin{cases}|x-y+3|\ge0\\2015.\left(2y-3\right)^{2016}\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\\left(2y-3\right)^{2016}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\2y-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\2y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\y=\frac{3}{2}\end{cases}}\)

Bạn thay vào tìm x

Mik cũng hok Toán 2

Hình như bn viết sai đề,là 1/x.(x+1) chứ

ukm mik xin lỗi mik viết sai đề đó

Đề sai nhé phải là x(x+1)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\Leftrightarrow\frac{x}{x+1}=\frac{2015}{2016}\Rightarrow x=2015\)

Vậy \(x=2015\)

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

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

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

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

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

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

<=> \(x=2017\)

Vậy x = 2017

đúng thì

Ta có: \(x^2\ge0;\left|x+y\right|\ge0;\forall x,y\)

=> \(M=2015+3\left(x^2+1\right)^{2016}+\left|x+y\right|^{2017}\)

\(\ge2015+3\left(0+1\right)^{2016}+0^{2017}=2018\)

Dấu "=" xảy ra khi và chỉ khi: \(\hept{\begin{cases}x^2=0\\\left|x+y\right|=0\end{cases}\Leftrightarrow x=y=0}\)

Vậy gtnn của M = 2018 đạt tại x = y = 0.

(x+2/2014)+1 + (x+1/2015)+1 = (x+2016)+1 + (x-1/2017)+1

(x+2016/2014) + (x+2016/2015) - (x+2016/2016) - (x-2016/2017)=0

=>(x+2016)(1/2014+1/2015-1/2016-1/2017)

vì 1/2014+1/2015-1/2016-1/2017 luôn khác 0 => x+2016=0

=> x=-2016