CMR:1 +( 1/1x 3 )+ 1 + (1/2 x 4) + 1 + (1/3 x 5) + ...... + 1+(1 /2017 x 2019) < 2
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31 tháng 8 2022
a: =>x-2017=0 và y-2018=0
=>x=2017; y=2018
b: =>3x-y=0 và y+2/3=0
=>y=-2/3 và 3x=-2/3
=>x=-2/9 và y=-2/3
c: =>3/4x-1/2=0 và 4/5y+6/25=0
=>x=2/3 và y=-3/10
DH
31 tháng 8 2019
a, \(\left(x+1\right)^2=169\)
\(\left(x+1\right)^2=13^2\)
\(x+1=13\)
\(x=13-1\)
\(x=12\)
VM
31 tháng 8 2019
1.
a) \(\left(x+1\right)^2=169\)
⇒ \(x+1=\pm13\)
⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{12;-14\right\}.\)
b) \(\left(x+3\right)^3=-\frac{1}{27}\)
⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)
⇒ \(x+3=-\frac{1}{3}\)
⇒ \(x=\left(-\frac{1}{3}\right)-3\)
⇒ \(x=-\frac{10}{3}\)
Vậy \(x=-\frac{10}{3}.\)
c) \(\left(2x-4\right)^4=\frac{1}{625}\)
⇒ \(2x-4=\pm\frac{1}{5}\)
⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)
Còn câu d) bạn làm tương tự như mấy câu trên.
Chúc bạn học tốt!
1 tháng 6 2019
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}\)
\(A=\frac{2019}{2020}\)
1 tháng 6 2019
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=1-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)
17 tháng 3 2019
Bài 2:
Đặt \(2017-x=a;2019-x=b;2x-4036=c\)
\(\Rightarrow a+b+c=0\)
Do \(a+b+c=0\Rightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
Có : \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab.\left(-c\right)+c^3=3abc\)
Do \(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)
\(\Rightarrow3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\2x-4036=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)
NN
13 tháng 8 2023
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~