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TT
0
SH
0
9 tháng 9 2019
=1/1.2+5/2.3+11/3.4+19/4.5+29/5.6+41/6.7
=1-1/2+5/2-5/3+11/3-11/4+19/4-19/5+29/5-29/6+41/6-41/7
=3+2+2+2+2-41/7
=77/7-41/7
=36/7
k nhé
9 tháng 9 2019
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=6-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\right)\)
\(=6-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=6-\left(1-\frac{1}{7}\right)=6-\frac{6}{7}=\frac{36}{7}\)
24 tháng 10 2021
1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=1+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}\)
TD
2
G
10 tháng 8 2014
a = 1/(1.2) + 5/(2.3) + ... + 89/(9.10)
a = [1-1/(1.2)] + [1-1/(2.3)] + ... + [1-1/(9.10)]
\(a=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{9.10}\right)\)
\(a=9-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right]\)
Ta có:
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
....
\(\frac{1}{9.10}=\frac{1}{9}-\frac{1}{10}\)
Cộng các vế ở trên lại:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
Vậy:
a = 9 - 9/10 = 81/10
VK
0
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
9 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
ta có: \(\frac{11}{12}.\left(\frac{19}{41}+\frac{22}{41}\right)\)
= \(\frac{11}{12}\)
\(\frac{11}{12}.\frac{19}{41}+\frac{11}{12}.\frac{22}{41}\)
\(=\frac{11}{12}.\left(\frac{19}{41}+\frac{22}{41}\right)\)
\(=\frac{11}{12}.1=\frac{11}{12}\)