cho mik hỏi 5 x 9 = ??? giúp mik nha
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Bài 1:
ta có: A = 11^9+11^8+..+11+1
=> 11A = 11^10+11^9+...+11^2+11
=> 11A-A = 11^10-1
10A = 11^10 -1
mà (11^10)-1 = (...1) - 1 = (...0) chia hết cho 10
=> A = (11^10-1):10 sẽ chia hết
=> A chia hết cho 5
Bài 2:
ta
Bài 1:
b) Ta có: \(D=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)
\(=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot0\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)
=0
\(72-3\left|x\right|=9\)
\(3\left|x\right|=72-9=63\)
\(\left|x\right|=63:3=21\)
\(\Rightarrow x=\pm21\)
\(-\frac{9}{46}-4\frac{1}{23}:\left(3\frac{1}{4}-x:\frac{3}{5}\right)+2\frac{8}{23}=1\)
=> \(-\frac{9}{46}-\frac{93}{23}:\left(\frac{13}{4}-x:\frac{3}{5}\right)+\frac{54}{23}=1\)
=> \(-\frac{9}{46}-\frac{93}{23}:\left(\frac{13}{4}-x:\frac{3}{5}\right)=1-\frac{54}{23}\)
=> \(-\frac{9}{46}-\frac{93}{23}:\left(\frac{13}{4}-x:\frac{3}{5}\right)=-\frac{31}{23}\)
=> \(\frac{93}{23}:\left(\frac{13}{4}-x:\frac{3}{5}\right)=-\frac{9}{46}-\left(-\frac{31}{23}\right)\)
=> \(\frac{93}{23}:\left(\frac{13}{4}-x:\frac{3}{5}\right)=-\frac{9}{46}+\frac{31}{23}=\frac{53}{46}\)
=> \(\frac{13}{4}-x:\frac{3}{5}=\frac{93}{23}:\frac{53}{46}\)
=> \(\frac{13}{4}-x:\frac{3}{5}=\frac{93}{23}\cdot\frac{46}{53}=\frac{186}{53}\)
=> \(x:\frac{3}{5}=\frac{13}{4}-\frac{186}{53}=-\frac{55}{212}\)
=> \(x=-\frac{55}{212}\cdot\frac{3}{5}=-\frac{33}{212}\)
Vậy : ....
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
Bằng 45