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\(S=2\cdot10+2\cdot12+2\cdot14+...+2\cdot20\)
\(S=2\cdot\left(10+12+14+16+18+20\right)\)
\(S=2\cdot\left[\left(20+10\right)\cdot6:2\right]\)
\(S=2\cdot90=180\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2.10+2.12+2.14+...+2.20\\ S=2.\left(10+12+14+...+20\right)\\ S=2.\left\{\left[\left(20-10\right):2+1\right].\left(20+10\right):2\right\}\\ S=2.\left\{\left[10:2+1\right].30:2\right\}\\ S=2.\left\{\left[5+1\right].30:2\right\}\\ S=2.\left\{6.30:2\right\}\\ S=2.\left\{180:2\right\}\\ S=2.90\\ S=180\)
TK :
S=2.10+2.12+2.14+...+2.20
S=2.[(10+20)+(12+18)+(14+16)]
S=2.(30+30+30)
S=2.90
S=180
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{8.2+2.14+2.90+2.6+2.12}{2.3.4+2.8+2.79+2.11}\)
\(=\frac{2\left(8+14+90+6+12\right)}{2\left(3.4+8+79+11\right)}\)
\(=\frac{130}{110}\)
\(=\frac{13}{11}\)
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Giải:
a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\)
\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\)
\(=\dfrac{1}{2^6}=\dfrac{1}{64}\)
b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\)
\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\)
\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\)
\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\)
\(=2\)
c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\)
\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\)
\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\)
\(=\dfrac{2.3}{3.5}\)
\(=\dfrac{2}{5}\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)
b. \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}=\dfrac{15^2.16^3\left(16-15\right)}{12^2.20^2\left(20-12\right)}=\dfrac{15^2.16^3}{12^2.20^2.8}=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.2^3}=\dfrac{3^2.5^2.2^{12}}{3^2.2^4.2^4.5^2.2^3}=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}=\dfrac{1}{2}\)
\(a\))\(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\)
\(=\dfrac{5^3.3^2.10.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\Leftrightarrow\dfrac{1.1.10.1}{5.1.2^7}=\dfrac{1}{64}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2.3.\left(1+2.3+7.7\right)}{3.5.\left(1+2.3+7.7\right)}=\frac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}=\frac{3^25^2.4^3.4^3-3^3.5^3.4^3.4^3}{3^2.4^2.4^3.5^3-4^2.5^2.3^3.4^3}\)
\(=\frac{3^2.5^2.4^3.4^3\left(4.4-3.5\right)}{3^2.5^2.4^3.4^2\left(5-4\right)}\)\(=\frac{4.1}{1}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{5^3.2.5.3^2.2^6}{5^4.3^2.2^{13}}=\dfrac{5^4.2^7.3^2}{5^4.3^2.2^{13}}=\dfrac{2^7}{2^{13}}=\dfrac{1}{64}\)
b) \(\dfrac{15^5.16^7}{12^5.20^5}=\dfrac{5^5.3^5.2^{28}}{2^{10}.3^5.2^{10}.5^5}=\dfrac{5^5.3^5.2^{28}}{2^{20}.3^5.5^5}=\dfrac{2^{28}}{2^{20}}=256\)
\(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}=\dfrac{5^3.2.5.3^2.2^6}{5^4.3^2.2^{13}}=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)
\(\dfrac{15^2.16^4.15^3.16^3}{12^2.20^3.20^2.12^3}=\dfrac{\left(3.5\right)^2.\left(2^4\right)^4.\left(3.5\right)^3.\left(2^4\right)^3}{\left(3.2^2\right)^2.\left(2^2.5\right)^3.\left(2^2.5\right)^2.\left(2^2.3\right)^3}=\dfrac{3^2.5^2.2^{16}.3^3.5^3.2^{12}}{3^2.2^4.2^6.5^3.2^4.5^2.2^6.3^3}=\dfrac{3^5.5^5.2^{28}}{3^5.5^5.2^{20}}=2^8=256\)
\(S=2\cdot10+2\cdot12+2\cdot14+\dots+2\cdot20\\=2\cdot(10+12+14+\dots+20)\)
Đặt \(A=10+12+14+\dots+20\)
Số các số hạng của $A$ là:
$(20-10):2+1=6$ (số)
Tổng $A$ bằng:
$(20+10)\cdot6:2=90$
Thay $A=90$ vào $S$, ta được:
$S=2\cdot90=180$
S=300