(x+2016)^2=-12.(-3)

(x+2016)^2=36

x+2016=6 hoặc x+2016=-6

x=-2010 ,x=-2022

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\) nên x+1=0

=>x=0-1

=>x-1

a:x+1/10+x+1/11+x+1/12=x+1/13+x+1/14

 <=>(x+1)(1/10 + 1/11+1/12) =(x+1)(1/13 + 1/14) 
<=>(x+1)(1/10 + 1/11+1/12 -1/13 -1/14)=0 
<=> x+1=0(vì biểu thức 1/10 + 1/11 +1/12-1/13-1/14#0) 
<=>x= -1

b:hình như sai đề

trả lời

nhân chéo lên là làm đc nha bn

hok tốt

\(\frac{x+2016}{-3}=-\frac{12}{x+2016}\)

\(\Rightarrow\left(x+2016\right)^2=-3.\left(-12\right)\)

\(\Rightarrow\left(x+2016\right)^2=36\)

\(\Rightarrow\left(x+2016\right)^2=6^2\)

\(\Rightarrow x+2016=6\)

\(\Rightarrow x=6-2016\)

\(\Rightarrow x=-2010\)

\(\frac{x+2016}{-3}=\frac{-12}{x+2016}\)

\(\Leftrightarrow(x+2016)\cdot(x+2016)=(-12)\cdot(-3)\)

\(\Leftrightarrow(x+2016)^2=36\)

\(\Leftrightarrow(x+2016)^2=6^2\)

\(\Leftrightarrow x+2016=\pm6\)

\(\Leftrightarrow\hept{\begin{cases}x+2016=6\\x+2016=-6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2010\\x=-2022\end{cases}}\)

Do đó,\(x\in\left\{-2010;-2022\right\}\)

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{...}

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

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

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

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

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

(2016.x+3.y+1).(2016^x+2016.x+2016.x+y)=225

=2016+3+1.(x+y).........

...................

tk đi chỉ tiếp

x=( 2^47+4^50)( 3^15-3^12)( 2^4-4^2)/2016

=(2^47+4^50)(3^15-3^12)(16-16)/2016

=0/2016=0 

Vậy x=0

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