Bài 1: Tìm x biết: \(\frac{x+5}{2014}+\frac{x+4}{2015}=\frac{x+3}{2016}+\frac{x+2}{2017}\)
Bài 2: Tìm cặp số nguyên (x;y) thoả mãn: \(\left|5x\right|+\left|2y+3\right|=7\)
Tìm x,biết:
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
=> x+2018=0
=> x=-2018
Tìm x biết:
\(\frac{x-1}{2017}+\frac{x-2}{2016}-\frac{x-3}{2015}=\frac{x-4}{2014}\)
Tìm x biết
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
b) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
c) \(\frac{x-1}{2017}+\frac{x-2}{2016}=\frac{x-3}{2015}+\frac{x-4}{2014}\)
d) \(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4029}{2014}+\frac{4030}{2015}\)
Tìm x biết:
\(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)
trừ mỗi vế cho 2 rồi tách -2 thành -1và -1
\(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)
\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
\(\Leftrightarrow\)\(x-1=0\) ( do 1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )
\(\Leftrightarrow\) \(x=1\)
CÂU 2:
a) Tìm x biết: \(\frac{x+2}{2017}+\frac{x+3}{2016}+\frac{x+4}{2015}+\frac{x+5}{1007}+\frac{x+2074}{11}=0\)
b) Tìm các cặp số nguyên (x;y) sao cho \(2019^{||x^2-y|-8|+y^2-1}=1\)
HELP ME, TICK TICK
tìm x biết \(\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2015}\right)\times x+2015=\frac{2016}{1}+\frac{2017}{2}+.....+\frac{4029}{2014}+\frac{4030}{2015}\)
Các bn ơi giúp mk vs:
Bài 1 tìm x biết
\(\frac{x+2015}{2016}\)+ \(\frac{x+2016}{2015}\)+\(\frac{x+2017}{2014}\)=-3
(x+\(\frac{1}{4}\))(x-\(\frac{2}{5}\))(x+\(\frac{4}{3}\))>0
bài 2Tìm GTLN của biểu thức sau
C=\(\frac{7x-8}{2x-3}\)
Bài 3 Tìm GTNN của biểu thức sau
a/ A=5-3(2x+1)^2
b/ B=\(\frac{1}{2\left(x-1\right)^2+3}\)
c/ C=\(\frac{^{x^2}+8}{x^2+2}\)
Bài hơi dài 1 chút các bn thông cảm cho mk nha ai giải đc bài nào thì xin hãy giúp mk nha! Mk cảm ơn trước
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
Tìm x :
\(\frac{x=2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)
\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)
\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
\(\Rightarrow x+4031=0\)
\(\Rightarrow x=-4031\)