D= \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}\)
giúp mình nha !!!! thanks trước
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11 tháng 7 2021
a) Ta có: \(7\cdot\dfrac{3}{14}-\dfrac{1}{14}\)
\(=\dfrac{3}{2}-\dfrac{1}{14}\)
\(=\dfrac{21}{14}-\dfrac{1}{14}\)
\(=\dfrac{10}{7}\)
b) Ta có: \(\dfrac{3}{2}+\dfrac{7}{4}:\dfrac{5}{2}\)
\(=\dfrac{3}{2}+\dfrac{7}{4}\cdot\dfrac{2}{5}\)
\(=\dfrac{3}{2}+\dfrac{7}{10}\)
\(=\dfrac{15}{10}+\dfrac{7}{10}=\dfrac{22}{10}=\dfrac{11}{5}\)
AH
Akai Haruma
Giáo viên
11 tháng 7 2021
Lời giải:
\(7\times \frac{3}{14}-\frac{1}{14}=\frac{7\times 3}{14}-\frac{1}{14}=\frac{21}{14}-\frac{1}{14}=\frac{21-1}{14}=\frac{20}{14}=\frac{2\times 10}{2\times 7}=\frac{10}{7}\)
\(\frac{3}{2}+\frac{7}{4}:\frac{5}{2}=\frac{3}{2}+\frac{7}{4}\times \frac{2}{5}=\frac{3}{2}+\frac{7\times 2}{4\times 5}=\frac{3}{2}+\frac{7\times 2}{2\times 2\times 5}\)
\(=\frac{3}{2}+\frac{7}{2\times 5}=\frac{3\times 5}{2\times 5}+\frac{7}{2\times 5}=\frac{3\times 5+7}{2\times 5}=\frac{22}{2\times 5}=\frac{2\times 11}{2\times 5}=\frac{11}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2022
Đặt \(A=\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}\)
\(7A=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{99}}\)
\(\Rightarrow7A-A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)
\(\Rightarrow6A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)
\(\Rightarrow A=\dfrac{1}{6}\left(\dfrac{1}{7}-\dfrac{1}{7^{100}}\right)\)
OY
12 tháng 7 2021
a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
A=0+0+0+...+0+\(\dfrac{13}{15}\)
A=\(\dfrac{13}{15}\)
12 tháng 7 2021
b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)
BF
19 tháng 2 2023
`3/4 + 5/6 = 9/12 + 10/12 = 19/12`
`1/2 + 7/12 = 6/12 + 7/12 = 13/12`
`2/3 xx 3/4 = 2/4 = 1/2`
`7/4 : 2 = 7/4 xx 1/2 = 7/8`
VQ
19 tháng 2 2023
\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)
\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)
\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)
\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)
8 tháng 1 2021
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
ND
1
NH
30 tháng 3 2023
\(\dfrac{3}{x}+\dfrac{x}{x+1}+\dfrac{x-3}{x}=\dfrac{13}{7}\left(x\ne0;x\ne-1\right)\)
\(< =>\dfrac{3\cdot7\left(x+1\right)}{7x\left(x+1\right)}+\dfrac{7x\cdot x}{7x\left(x+1\right)}+\dfrac{7\left(x-3\right)\left(x+1\right)}{7x\left(x+1\right)}=\dfrac{13x\left(x+1\right)}{7x\left(x+1\right)}\)
suy ra
\(21x+21+7x^2+7\left(x^2+x-3x-3\right)=13x^2+13x\)
\(< =>21x+21+7x^2+7x^2+7x-21x-21=13x^2+13x\)
\(< =>7x^2+7x^2-13x^2+21x+7x-21x-13x+21-21=0\)
\(< =>x^2-6x=0\\ < =>x\left(x-6\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(loại\right)\\x=6\left(tm\right)\end{matrix}\right.\)
2D=1+1/7+1/72+...+1/799
2D-D=(1+1/7+1/72+...+1/799)-(1/7+1/72+...1/7100)
D=1-1/7100
Giải:
D=1/7+1/72+1/73+...+1/7100
7D=1+1/7+1/72+...+1/799
7D-D=(1+1/7+1/72+...+1/799)-(1/7+1/72+1/73+...+1/7100)
6D=1-1/7100
D=1-1/7100/6
Chúc bạn học tốt!