chứng minh:\(2^{27}+2^{25}\)chia hết cho 5
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+3\right)^2-2=23\)
\(\Rightarrow\left(2x+3\right)^2=23+2\)
\(\Rightarrow\left(2x+3\right)^2=25\)
\(\Rightarrow\left(2x+3\right)^2=5^2\)
\(\Rightarrow2x+3=5\)
\(\Rightarrow2x=5-3\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=2:2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{24.25}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{600}\right)=\frac{1}{2}.\frac{299}{600}=\frac{299}{1200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{c+a}{5}=\frac{b+c}{4}=\frac{a+b}{3}=\frac{c+b-b-c+a+b}{5-4+3}=\frac{2a}{4}=\frac{a}{4}\left(1\right)\)
Từ (1) có: \(\frac{b+c}{4}=\frac{a+b}{3}\Leftrightarrow3b+3c=4a+4b\Leftrightarrow b=3c-4a\left(2\right)\)
Thế 2 vào biểu thức M ta có: \(M=10a+3c-4a-7c+2017=6a-4c+2017\left(3\right)\)
Từ (1) có\(:\frac{c+a}{5}=\frac{a}{2}\Leftrightarrow2c+2a=5a\Leftrightarrow2c=3a\Leftrightarrow4c=6a\left(4\right)\)
Thế (4) vào (3) ta có: \(M=6a-6a+2017=2017\)
Vậy GT M = 2017
+ Ta có : \(\frac{a+b}{3}=\frac{b+c}{4}\Rightarrow4a+4b=3b+3c\)
\(\Rightarrow4a+b=3c\)
+ \(\frac{a+b}{3}=\frac{c+a}{5}\Rightarrow5a+5b=3c+3a\)
\(\Rightarrow2a+5b=3c\)
+ \(\frac{b+c}{4}=\frac{c+a}{5}\Rightarrow5b+5c=4c+4a\)
\(\Rightarrow5b+c=4a\)
+ Ta có : \(\hept{\begin{cases}4a+b=3c\\5b+3a=3c\end{cases}\Rightarrow4a+b=5b+2a}\)
\(\Rightarrow2a=4b\)
\(\Rightarrow a=2b\)
+ Ta có : \(4a+b=3c\)
\(\Rightarrow4.2b+b=3c\)
\(9b=3c\)
\(\Rightarrow3b=c\)
+ Ta có : \(M=10a+b-7c+2017\)
\(=10.2b+b-7.3b+2017\)
\(=20b+b-7.3b+2017\)
\(=21b-21b+2017\)
\(=0+2017=2017\)
Vậy M =2017
Chúc bạn học tốt !!!
Ta có : 227=23.(24)6=8.\(\left(\overline{...6}\right)\)=\(\overline{...8}\)
225=2.(24)6=2.\(\left(\overline{...6}\right)\)=\(\overline{...2}\)
\(\Rightarrow\)227+225=\(\left(\overline{...8}\right)+\left(\overline{...2}\right)=\overline{...0}⋮5\)
Vậy 227+225 \(⋮\)5.