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\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{22}\)
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\)=\(\frac{8}{22}=\frac{4}{11}\)
Hok tốt
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5.4
a: Xét ΔADB và ΔEDB có
BA=BE
\(\widehat{ABD}=\widehat{EBD}\)
BD chung
Do đó: ΔADB=ΔEDB
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2^3.3^5.6^2}{2^5.3^7}=\frac{2^3.3^5.2^2.3^2}{2^5.3^7}=\frac{2^5.3^7}{2^5.3^7}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\)
\(=\frac{3}{5}.\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(< \frac{3}{5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{100}\right)\)
\(< \frac{3}{5}.1=\frac{3}{5}=0,6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^5\cdot6^7}{8^5\cdot3^5}=\frac{2^5\cdot\left(2\cdot3\right)^7}{\left(2\cdot4\right)^5\cdot3^5}=\frac{2^5\cdot2^7\cdot3^7}{2^5\cdot4^5\cdot3^5}=\frac{2^7\cdot3}{4^5}\)
x; y tỉ lệ thuận với 2 và 5 nên :
\(\frac{x}{2}=\frac{y}{5}\)
\(\Rightarrow\frac{x-y}{2-5}=\frac{x}{2}=\frac{y}{5}\) mà x - y = 21
\(\Rightarrow\frac{21}{-3}=\frac{x}{2}=\frac{y}{5}\)
\(\Rightarrow-7=\frac{x}{2}=\frac{y}{5}\)
\(\Rightarrow\hept{\begin{cases}x=-7\cdot2=-14\\y=-7\cdot5=-35\end{cases}}\)
vậy_
x; y tỉ lệ nghịch với 2 và 5 nên :
2x = 5y
=> 2x/10 = 5y/10
=> x/5 = y/2
=> x/5 = 2y/4
=> x + 2y/5 + 4 = x/5 = y/2 mà x + 2y = 54
=> 54/9 = x/5 = y/2
=> 6 = x/6 = y/2
=> x = 6*6 = 36; y = 6*2 = 12
vậy_