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Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\Rightarrow x=\dfrac{1009}{9}\)
Vậy.............
Chúc bạn học tốt!!!
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\)
\(\Rightarrow18x=2018-0\)
\(\Rightarrow18x=2018\)
\(\Rightarrow x=2018:18\)
\(\Rightarrow x=\dfrac{1009}{9}\)
1,
đặt A= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\)
2A=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\)
2A-A=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))
A=1-\(\dfrac{1}{2017}\)
A=\(\dfrac{2016}{2017}\)
vậy A=\(\dfrac{2016}{2017}\)
Giải:
a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)
\(\Rightarrow x=\dfrac{12.-4}{16}=-3\)
\(\Rightarrow y=\dfrac{16.21}{12}=28\)
\(\Rightarrow z=\dfrac{12.80}{16}=60\)
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\) =0
\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\) \(=0+\dfrac{2}{5}\)
\(x.\dfrac{11}{15}\) \(=\dfrac{2}{5}\)
x \(=\dfrac{2}{5}:\dfrac{11}{15}\)
x \(=\dfrac{6}{11}\)
c) (2x-3)(6-2x)=0
⇒2x-3=0 hoặc 6-2x=0
x=3/2 hoặc x=3
d) \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\)
\(2x-5=\dfrac{-13}{2}\)
\(2x=\dfrac{-13}{2}+5\)
\(2x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:2\)
\(x=\dfrac{-3}{4}\)
e) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}:2\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\) hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\)
\(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)
Nhân cả 2 vế với 3 ta có:
\(pt\Leftrightarrow2x-\dfrac{6}{y}=1\Leftrightarrow2x=1+\dfrac{6}{y}\)
Nhận thấy rằng 2x là số nguyên, 1 là số nguyên nên \(\dfrac{6}{y}\) cũng là số nguyên
=> y ∈ Ư(6) = {\(\pm\)1; \(\pm\)2; \(\pm\)3; \(\pm\)6}
Mà 2x là số chẵn => \(1+\dfrac{6}{y}\) là số chẵn => y ∈ {\(\pm\)2; \(\pm\)6}
+) \(y=-6\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{-6}\right)=0\)
+) \(y=-2\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{-2}\right)=-1\)
+) \(y=2\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{2}\right)=2\)
+) \(y=6\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{6}\right)=1\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(2017>2016>2015>2014\) nên
\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\Rightarrow x=1009\)
Vậy...........
Chúc bạn học tốt!!!
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)
\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\)
\(\Rightarrow2x=2018-0\)
\(\Rightarrow2x=2018\)
\(\Rightarrow x=2018:2\)
\(\Rightarrow x=1009\)