Tìm x nếu |2016-x|+2016=x
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\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)
\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)
=> x = 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
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 \(x=75\)
\(\frac{x+5}{2017}+\frac{x+6}{2016}+\frac{x+7}{2015}=-3\)
\(\left(\frac{x+5}{2017}+1\right)+\left(\frac{x+6}{2016}+1\right)+\left(\frac{x+7}{2015}+1\right)=0\)
\(\frac{x+2022}{2017}+\frac{x+2022}{2016}+\frac{x+2022}{2015}=0\)
\(\left(x+2022\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(x+2022=0\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\ne0\right)\)
x=-2022