tìm ƯCLN của 56 và 140
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\(x+x:2+x:3=11\)
=>\(x+\dfrac{1}{2}x+\dfrac{1}{3}x=11\)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+1\right)=11\)
=>\(x\cdot\dfrac{11}{6}=11\)
=>\(x=11:\dfrac{11}{6}=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\dfrac{1}{18}+\dfrac{1}{18\cdot9}+\dfrac{1}{162\cdot9}+\dfrac{1}{1452\cdot9}\)
\(=\dfrac{1}{9}\left(\dfrac{1}{2}+\dfrac{1}{18}+\dfrac{1}{162}+\dfrac{1}{1452}\right)\)
\(=\dfrac{1}{9}\cdot\left(\dfrac{81}{162}+\dfrac{9}{162}+\dfrac{1}{162}+\dfrac{1}{1452}\right)\)
\(=\dfrac{1}{9}\cdot\left(\dfrac{91}{162}+\dfrac{1}{1452}\right)=\dfrac{1}{9}\cdot\left(\dfrac{22022}{39204}+\dfrac{27}{39204}\right)\)
\(=\dfrac{1}{9}\cdot\dfrac{22049}{39204}=\dfrac{22049}{352836}\)
\(S=\dfrac{1}{18}+\dfrac{1}{18\cdot9}+\dfrac{1}{162\cdot9}+\dfrac{1}{1452\cdot9}\\ =\dfrac{1}{9\cdot2}+\dfrac{1}{9^2\cdot2}+\dfrac{1}{9^3\cdot2}+\dfrac{1}{9^4\cdot2}\\ =\dfrac{1}{2}\left(\dfrac{1}{9}+\dfrac{1}{9^2}+\dfrac{1}{9^3}+\dfrac{1}{9^4}\right)\)
Đặt:
\(K=\dfrac{1}{9}+\dfrac{1}{9^2}+\dfrac{1}{9^3}+\dfrac{1}{9^4}=>9K=1+\dfrac{1}{9}+\dfrac{1}{9^2}+\dfrac{1}{9^3}\\ =>9K-K=1+\dfrac{1}{9}+\dfrac{1}{9^2}+\dfrac{1}{9^3}-\dfrac{1}{9}-\dfrac{1}{9^2}-\dfrac{1}{9^3}-\dfrac{1}{9^4}\\ =>8K=1-\dfrac{1}{9^4}=\dfrac{9^4-1}{9^4}\\ =>K=\dfrac{9^4-1}{8\cdot9^4}\)
\(=>S=\dfrac{1}{2}\cdot\dfrac{9^4-1}{8\cdot9^4}=\dfrac{9^4-1}{16\cdot9^4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(12-\left(-0,6\right)=12+0,6=12,6\)
b: \(\left(-0,25\right)\cdot1,24=-\dfrac{1}{4}\cdot\dfrac{31}{25}=\dfrac{-31}{100}\)
c: \(\left(-4,5\right)+\left(17,45\right)=17,45-4,5=12,95\)
d: \(\dfrac{-5,85}{2,25}=\dfrac{-585}{225}=-\dfrac{13}{5}=-2,6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 3:
\(u_1=\dfrac{2\cdot1+1}{1+2}=\dfrac{3}{3}=1\)
\(u_4=\dfrac{2\cdot4+1}{4+2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(u_5=\dfrac{2\cdot5+1}{5+2}=\dfrac{11}{7}\)
Câu 2:
\(u_n=u_1+\left(n-1\right)\cdot d\)
=>\(-3\left(n-1\right)+4=-41\)
=>-3(n-1)=-45
=>n-1=15
=>n=16
Câu 1:
Tổng của 50 số hạng đầu là 5150
=>\(\dfrac{n\cdot\left[2\cdot u_1+\left(n-1\right)\cdot d\right]}{2}=5150\)
=>\(\dfrac{50\left(2\cdot5+\left(50-1\right)\cdot d\right)}{2}=5150\)
=>\(25\left(10+49d\right)=5150\)
=>49d+10=206
=>49d=196
=>d=4
\(u_{10}=u_1+9d=5+9\cdot4=5+36=41\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2,5=\dfrac{5}{2}\)
Hiệu số phần bằng nhau là: 5-2=3(phần)
Số quả trứng của bà thứ nhất là:
24:3x5=40(quả)
Số quả trứng của bà thứ hai là:
40-24=16(quả)
Hiệu số phần bằng nhau là: 5-2=3(phần)
Số quả trứng của bà thứ nhất là:
24:3x5=40(quả)
Số quả trứng của bà thứ hai là:
40-24=16(quả)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{4\text{x}7}+\dfrac{5}{7\text{x}10}+...+\dfrac{5}{58\text{x}61}\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{3}{4\text{x}7}+\dfrac{3}{7\text{x}10}+...+\dfrac{3}{58\text{x}61}\right)\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{58}-\dfrac{1}{61}\right)\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{1}{4}-\dfrac{1}{61}\right)=\dfrac{5}{3}\text{x}\dfrac{57}{244}=\dfrac{95}{244}\)
\(\dfrac{5}{4\times7}+\dfrac{5}{7\times10}+\dfrac{5}{10\times13}+...+\dfrac{5}{58\times61}\\ =\dfrac{5}{3}\times\left(\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+...+\dfrac{3}{58\times61}\right)\\ =\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{58}-\dfrac{1}{61}\right)\\ =\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{61}\right)\\ =\dfrac{5}{3}\times\dfrac{57}{244}=\dfrac{95}{244}\)
\(56=2^3\cdot7;140=2^2\cdot5\cdot7\)
=>\(ƯCLN\left(56;140\right)=2^2\cdot7=28\)
Ta có:
\(56=7\cdot8=7\cdot2^3\)
\(140=4\cdot35=2^2\cdot5\cdot7\)
\(=>ƯCLN\left(56;140\right)=7\cdot2^2=28\)