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\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\right)\)
\(=3.\left(1-\frac{1}{46}\right)\)
\(=3.\frac{45}{46}\)
\(=\frac{135}{46}\)
~Học tốt~
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nhân chéo chia ngang
x= -7*36/12=-21
muốn chắc chắn thì thử lại
\(\frac{-7}{12}=\frac{21}{36}\)theo quy đồng cũng đúng
Viết lại là:
\(\frac{-7}{12}=\frac{x}{36}\)
=> \(-7.36=12.x\)
=> \(-252=12.x\)
=> \(x=\frac{-252}{12}\)
=> \(x=-21\)
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a; \(\dfrac{x-1}{12}\) = \(\dfrac{5}{3}\)
\(x-1\) = \(\dfrac{5}{3}\) \(\times\) 12
\(x\) - 1 = 20
\(x\) = 20 + 1
\(x\) = 21
b; \(\dfrac{-x}{8}\) = \(\dfrac{-50}{x}\)
-\(x\).\(x\) = -50.8
-\(x^2\) = -400
\(x^2\) = 400
\(\left[{}\begin{matrix}x=-20\\x=20\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-20; 20}
c; \(\dfrac{x}{3}\) = \(\dfrac{14}{x+1}\)
\(x\).(\(x\)+1) = 14.3
\(x^2\) + \(x\) = 42
\(x^2\) + \(x\) - 42 = 0
\(x^2\) - 6\(x\) + 7\(x\) - 42 = 0
\(x\).(\(x\) - 6) + 7.(\(x\) - 6) = 0
(\(x\) - 6).(\(x\) + 7) = 0
\(\left[{}\begin{matrix}x-6=0\\x+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-7; 6}
d; \(x-\dfrac{2}{9}\) = \(\dfrac{1}{6}\)
\(x\) = \(\dfrac{1}{6}\) + \(\dfrac{2}{9}\)
\(x\) = \(\dfrac{7}{18}\)
Vậy \(x\) = \(\dfrac{7}{18}\)
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125.(-8).(-25).9.4.1002:3
=125.(-8).(-25).4.1002.9:3
=(-1000).(-100).10000.3
=3000000000
Chúc bn học tốt
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Ta có : \(A=3+3^2+3^3+3^4+...+3^{25}\)
\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{23}+3^{24}+3^{25}\right)\)
\(=3+3\left(3+3^2+3^3\right)+...+3^{22}\left(3+3^2+3^3\right)\)
\(=3+3.39+...+3^{22}.39\)
\(=3+39\left(3+...+3^{22}\right)\)
\(\Rightarrow A\)chia cho 39 dư 3
\(\Rightarrow A\)không chia hết cho 39 ( đpcm )
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minh giai y 1con y 2 neu can giup minh se giai cho
=(-1+3)+(-5+7)+....+(97-99)=-2+-2+...+-2(co 25 so 2)=-2.25=-50
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Ta có:
+) \(\frac{2013.2012-1}{2013.2012}=1-\frac{1}{2013.2012}\)
+) \(\frac{2012.2011-1}{2012.2011}=1-\frac{1}{2012.2011}\)
Vì \(\frac{1}{2013.2012}< \frac{1}{2012.2011}\Rightarrow1-\frac{1}{2013.2012}>1-\frac{1}{2012.2011}\)
Vậy \(\frac{2013.2012-1}{2013.2012}>\frac{2012.2011-1}{2012.2011}\)
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96 - 3(x+1 ) = 42
=> 3(x+1) = 96 - 42 = 54
=> x + 1 = 54 : 3 = 18
x + 1 = 18
=> x = 17
96 - 3 ( x + 1 ) = 42
3 ( x + 1 ) = 96 - 42
3 (x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 - 1
x = 17
a: \(\dfrac{3}{5}+3\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):7\dfrac{2}{3}\)
\(=\dfrac{3}{5}+\dfrac{23}{6}\left(11+\dfrac{5}{20}-9-\dfrac{1}{4}\right):\dfrac{23}{3}\)
\(=\dfrac{3}{5}+\dfrac{23}{6}\cdot2\cdot\dfrac{3}{23}=\dfrac{3}{5}+1=\dfrac{8}{5}\)
b: \(\dfrac{7}{13}\cdot\dfrac{7}{15}-\dfrac{1}{3}\cdot\dfrac{21}{39}+\dfrac{49}{91}\cdot\dfrac{8}{15}\)
\(=\dfrac{7}{13}\cdot\dfrac{7}{15}-\dfrac{1}{3}\cdot\dfrac{7}{13}+\dfrac{7}{13}\cdot\dfrac{8}{15}\)
\(=\dfrac{7}{13}\left(\dfrac{7}{15}+\dfrac{8}{15}-\dfrac{1}{3}\right)=\dfrac{7}{13}\cdot\dfrac{2}{3}=\dfrac{14}{39}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}\)
\(=\dfrac{1}{2023}\)
d: \(\dfrac{2}{4\cdot5}+\dfrac{2}{5\cdot6}+...+\dfrac{2}{99\cdot100}\)
\(=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{100}\right)=2\cdot\dfrac{24}{100}=\dfrac{48}{100}=\dfrac{12}{25}\)
e: \(\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+...+\dfrac{3}{99\cdot101}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
f: \(\dfrac{10}{3\cdot6}+\dfrac{10}{6\cdot9}+...+\dfrac{10}{96\cdot99}\)
\(=\dfrac{10}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{96\cdot99}\right)\)
\(=\dfrac{10}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{96}-\dfrac{1}{99}\right)\)
\(=\dfrac{10}{3}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=\dfrac{10}{3}\cdot\dfrac{32}{99}=\dfrac{320}{297}\)