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16 tháng 11 2016

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

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

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

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

\(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)

=> x = 2017

26 tháng 9 2017

      B1 a A = 2/3+1/6-1/2=5/6-1/2=2/6=1/3

   b B=3.{5.[(25+8):11]-16}+2015=3.{5.[33:11]-16}=3.{5.3-16}+2015

      =3.{15-16}+2015=3.(-1)+2015=-3+2015=2012

      B2     8.6+288 :(x-3)2=50

                8.6+288:(x-3)=50:2

               8.6+288:(x-3)=25

               288:(x-3)=25-8.6

               288:(x-3)=-23

               x-3=-23.288

              X-3=-6624

              x=-6624+3

              X=-6627

   

26 tháng 9 2017

bai 1:  a) \(A=\frac{2}{3}+\frac{5}{6}:5-\frac{1}{18}.\left(-3\right)^2\)

 \(A=\frac{2}{3}+\frac{1}{6}-\frac{1}{18}.9\)

\(A=\frac{4}{6}+\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{5}{6}-\frac{3}{6}\)

\(A=\frac{2}{6}=\frac{1}{3}\)

b) \(B=3\left\{5.\left[\left(5^2+2^3\right):11\right]-16\right\}+2015\)

\(B=3\left\{5.\left[\left(25+8\right):11\right]-16\right\}+2015\)

\(B=3\left\{5.\left[33:11\right]-16\right\}+2015\)

\(B=3\left\{5.3-16\right\}+2015\)

\(B=3\left\{15-16\right\}+2015\)

\(B=3.\left(-1\right)+2015\)

\(B=-3+2015\)

\(B=2012\)

bai 2:  \(6.8+288:\left(x-3\right).2=50\)

\(48+288:\left(x-3\right).2=50\)

\(288:\left(x-3\right).2=50-48\)

\(288:\left(x-3\right).2=2\)

\(\left(x-3\right).2=288:2\)

\(\left(x-3\right).2=144\)

\(x-3=144:2\)

\(x-3=72\)

\(x=75\)

vay \(x=75\)

28 tháng 9 2016

Ấy nhầm đề

10 tháng 7 2016

\(\frac{x+5}{2017}+\frac{x+6}{2016}+\frac{x+7}{2015}=-3\)

\(\left(\frac{x+5}{2017}+1\right)+\left(\frac{x+6}{2016}+1\right)+\left(\frac{x+7}{2015}+1\right)=0\)

\(\frac{x+2022}{2017}+\frac{x+2022}{2016}+\frac{x+2022}{2015}=0\)

\(\left(x+2022\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)

\(x+2022=0\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\ne0\right)\)

x=-2022