3x+1/2018 + 3x+2/2017 = 3x+3/2016 + 3x+4/2015
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DD
0
TN
Tìm \(x\varepsilonℝ\) biết \(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)
2
KY
10 tháng 5 2018
\(\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}=\dfrac{3x+12}{2015}\)
\(\Rightarrow\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}-\dfrac{3x+12}{2015}=0\)
\(\Rightarrow\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}+\dfrac{3\cdot\left(x+4\right)}{2015}=0\)
\(\Rightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)+\left(\dfrac{3}{2015}\cdot\left(\dfrac{x+4}{2015}+1\right)\right)=0\)
\(\Rightarrow\left(x+2019\right)\cdot\left(\dfrac{1}{2018}+\dfrac{1}{2017}+\dfrac{1}{2016}+\left(\dfrac{3}{2015}\cdot\dfrac{1}{2005}\right)\right)=0\)
\(\Rightarrow x+2019=0\\ \Rightarrow x=-2019\)
BB
1
15 tháng 7 2022
\(f\left(x\right)=x^3-3x^2+3x-1+4=\left(x-1\right)^3+4\)
Lấy x1,x2 thuộc R sao cho x1<x2
\(A=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\left(x_1-1\right)^3-\left(x_2-1\right)^3}{x_1-x_2}\)
\(=\dfrac{\left(x_1-1-x_2+1\right)\left[\left(x_1-1\right)^2+\left(x_1-1\right)\left(x_2-1\right)+\left(x_2-1\right)^2\right]}{x_1-x_2}\)
\(=\left(x_1-1\right)^2+\left(x_1-1\right)\left(x_2-1\right)+\left(x_2-1\right)^2>0\)
=>A>0
Do đó: Hàm số đồng biến với x thuộc R
Do đó: \(f\left(\dfrac{2018}{2017}\right)< f\left(\dfrac{2017}{2016}\right)\)
CC
Tìm \(x\varepsilonℝ\) biết \(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)
6
YY
10 tháng 5 2018
\(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{3x+12}{2015}+3\)
\(\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{3 \left(x+2019\right)}{2015}\)
\(\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\right)=0\)
mà \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\ne0\Rightarrow x+2019=0\Leftrightarrow x=-2019\)
WH
11 tháng 4 2018
Nguyễn Tiến Đạt
a)\(|3x-5|=|x+2|\)
=> Ta có 2 trường hợp
*) TH1: 3x-5=x+2
=>3x-x=2+5
=>2x=7
=>x=7:2\(\Rightarrow x=\frac{7}{2}\)
*)TH2: -3x+5=x+2
\(\Rightarrow5-3x=x+2\)
\(\Rightarrow5-2=x+3x\)
\(\Rightarrow3=4x\)
\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)
23 tháng 5 2022
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
9 tháng 10 2016
Ta sẽ xét tính biến thiên của hàm số :
Ta có \(f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4\)
\(f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3\)
\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]\)
\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)< 0\)
\(\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)< 0\Rightarrow f\left(\frac{2017}{2016}\right)< f\left(\frac{2016}{2015}\right)\)
20 tháng 9 2019
Ta sẽ xét tính biến thiên của hàm số :
Ta có f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4f(x)=(x3−3x2+3x−1)+4=(x−1)3+4
f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3f(20162017)−f(20152016)=(20162017−1)3−(20152016−1)3
=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]=(20161−20151)[(20162017−1)2+(20152016−1)2+(20162017−1)(20152016−1)]
=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)< 0=(20161−20151)(201621+201521+20161.20151)<0
\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)< 0\Rightarrow f\left(\frac{2017}{2016}\right)< f\left(\frac{2016}{2015}\right)⇒f(20162017)−f(20152016)<0⇒f(20162017)<f(20152016)
\(\dfrac{3x+1}{2018}+\dfrac{3x+2}{2017}=\dfrac{3x+3}{2016}+\dfrac{3x+4}{2015}\)
\(\Leftrightarrow\left(\dfrac{3x+1}{2018}+1\right)+\left(\dfrac{3x+2}{2017}+1\right)=\left(\dfrac{3x+3}{2016}+1\right)+\left(\dfrac{3x+4}{2015}+1\right)\)
\(\Leftrightarrow\dfrac{3x+2019}{2018}+\dfrac{3x+2019}{2017}-\dfrac{3x+2019}{2016}-\dfrac{3x+2019}{2015}=0\)
\(\Leftrightarrow\left(3x+2019\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)
Mà \(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}< 0\)
\(\Rightarrow-\left(3x+2019\right)=0\Leftrightarrow x=-673\)