TÍNH
C = 7 + 7^2 + 7^3 + ... + 7^99
K
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H
8 tháng 5 2023
\(1,\\ c,\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\dfrac{17\times7-3\times6-5\times7}{6\times7}=\dfrac{119-18-35}{42}=\dfrac{66}{42}=\dfrac{11}{7}\\ 2,\\ a,\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\\ =\dfrac{6:3}{9:3}+\dfrac{7+3}{15}\\ =\dfrac{2}{3}+\dfrac{10}{15}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{2+2}{3}=\dfrac{4}{3}\)
\(b,\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\\ =\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\\ =\dfrac{7}{2}+\dfrac{29-11}{12}\\ =\dfrac{7}{2}+\dfrac{18}{12}\\ =\dfrac{7}{2}+\dfrac{18:6}{12:6}\\ =\dfrac{7}{2}+\dfrac{3}{2}\\ =\dfrac{7+3}{2}=\dfrac{10}{2}=5\)
\(c,\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\\ =\dfrac{26}{25}-\dfrac{3\times5}{5\times5}-\dfrac{2\times5}{5\times5}\\ =\dfrac{26-15-10}{25}\\ =\dfrac{1}{25}\)
S
8 tháng 5 2023
Bài 1:
\(\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\left(\dfrac{17}{6}-\dfrac{5}{6}\right)-\dfrac{3}{7}=2-\dfrac{3}{7}=\dfrac{14}{7}-\dfrac{3}{7}=\dfrac{11}{7}\)
Bài 2:
a/\(\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\)
\(=\dfrac{2}{3}+\left(\dfrac{3}{15}+\dfrac{7}{15}\right)\)
\(=\dfrac{2}{3}+\dfrac{2}{3}\)
\(=\dfrac{4}{3}\)
b/\(\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\)
\(=\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\)
\(=\dfrac{7}{2}+\dfrac{3}{2}\)
\(=5\)
c/\(\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\)
\(=\dfrac{26}{25}-\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{26}{25}-1\)
\(=\dfrac{26}{25}-\dfrac{25}{25}\)
\(=\dfrac{1}{25}\)
#Đang Bận Thở
23 tháng 7 2023
\(=\left[\dfrac{9-3}{162}\cdot\dfrac{81}{17}+\dfrac{35}{34}\right]:\left(\dfrac{3}{5}+\dfrac{7}{102}\right)\cdot\dfrac{102}{5}+2017\)
\(=\left[\dfrac{6}{2}\cdot\dfrac{1}{17}+\dfrac{35}{34}\right]:\dfrac{341}{510}\cdot\dfrac{102}{5}+2017\)
\(=\dfrac{41}{34}\cdot\dfrac{510}{341}\cdot\dfrac{102}{5}+2017=\dfrac{12546}{341}+2017=\dfrac{700343}{341}\)
NT
0
WH
7 tháng 5 2018
Đặt \(A=1+7^2+7^3+7^4+...+7^{99}\)
\(\Rightarrow7A=7\left(1+7^2+7^3+7^4+...+7^{99}\right)\)
\(\Rightarrow7A=7+7^3+7^4+7^5+...+7^{100}\)
\(\Rightarrow7A-A=\left(7+7^3+7^4+7^5+...+7^{100}\right)-\left(1+7^2+7^3+7^4+...+7^{99}\right)\)
\(\Rightarrow6A=7^{100}-1\)
\(\Rightarrow A=\frac{7^{100}-1}{6}\)
=> 7C= 72+73+.....+7100
=> 7C-C =6C=7100-7
=> C=(7100-7):6
study well
k nha
7C=72+73+74+...+7100
->7C-C=6C
->=6C
=7100-7
->C=(7100 -7):6