HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(BC\left(68,85,170\right)< 2000=\left\{340,680,1020,1360,1700\right\}\)
C1: \(\left(x-1\right)+\left(x-20\right)+...+\left(x-10\right)=165\)
\(\left(x+x+...+x\right)-\left(1+2+...+10\right)=165\)
\(\left(x.10\right)+\left(1+10\right).10:2=165\)
\(\left(x.10\right)+55=165\)
\(x.10=165-55\)
\(x.10=110\)
\(x=110:10\)
\(x=11\)
Vậy \(x=11\)
C2: \(\left(x-1\right)+\left(x-20\right)+...+\left(x-10\right)=165\)
\(\Rightarrow\) \(10x-\left(1+2+...+10\right)=165\)
\(\Rightarrow\) \(10x+55=165\)
\(\Rightarrow\) \(10x=165-55\)
\(\Rightarrow\) \(10x=110\)
\(\Rightarrow\) \(x=110:10\)
\(\Rightarrow\) \(x=11\)
Ta có:
\(648=2^3.3^4\)
\(540=2^2.3^3.5\)
\(ƯCLN\left(648,540\right)=2^2.3^3=108\)
\(ƯC\left(648,540\right)=\left\{2,3,4,6,9,12,18,27,36,54,108\right\}\)
\(ƯCLN\left(648,540\right)=\left\{108\right\}\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{512}-\dfrac{1}{512}\)
\(=1-\dfrac{1}{512}\)
\(=\dfrac{511}{512}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)
\(160-\left(2^3.5^2-6.25\right)\)
\(=160-\left(8.25-6.25\right)\)
\(=160-\left[25.\left(8-6\right)\right]\)
\(=160-\left[25.2\right]\)
\(=160-50\)
\(=110\)
\(=160-100\)
\(=60\)
\(\left|2x+1\right|=\dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}2x+1=\dfrac{3}{5}\\2x+1=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{5}-1\\2x=-\dfrac{3}{5}-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=-\dfrac{2}{5}\\2x=-\dfrac{8}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}:2\\x=-\dfrac{8}{5}:2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{5};-\dfrac{4}{5}\right\}\)
a) \(135+\left(-2\right)\cdot14-135+28\)
\(=135+\left(-28\right)-135+28\)
\(=\left(135-135\right)+\left[\left(-28\right)+28\right]\)
\(=0+0\)
\(=0\)
b) \(-18\cdot2+18\cdot\left(-6\right)+\left(-18\right)\cdot3\)
\(=-18\cdot2+\left(-18\right).6+\left(-18\right).3\)
\(=-18\cdot\left(2+6+3\right)\)
\(=-18\cdot11\)
\(=-198\)