Tính tổng : 1+7 + 72+73+ ....+ 72015
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\(A=7+7+7^2+...+7^{100}\)
\(7A=7^2+7^2+7^3+...+7^{101}\)
\(A=14+7^2+7^{101}\)
\(S=7+7^2+7^3+...7^{20}\)
Ta có: \(7S=7.\left(7+7^2+7^3+...+7^{20}\right)\)
\(7S=7^2+7^3+7^4+...+7^{21}\)
\(7S-S=\left(7^2+7^3+7^4+...+7^{21}\right)-\left(7+7^2+7^3+...+7^{20}\right)\)
\(6S=\left(7^{21}-7\right)\)
\(S=\left(7^{21}-7\right):6\)
Chúc bạn học tốt
A = 1 + 1/110 + 1 + 1/90 + ... + 1 + 1 /2
A = 10 + 1/1.2+ 1 /2.3 + ... + 1/9.10 + 1/10.11
A = 10 + 1/1 - 1/2 + 1 /2 - 1/3 + ... + 1/9 - 1/10 + 1/10 - 1/11
A = 10 + 1/1 - 1/11
A = 10 + 10/11
A = 120/11
A = \(\frac{111}{110}+\frac{91}{90}+\frac{73}{72}+...+\frac{13}{12}+\frac{7}{6}+\frac{3}{2}\)
A = \(\left(\frac{1}{2}+1\right)+\left(\frac{1}{6}+1\right)+\left(\frac{1}{12}+1\right)+....+\left(\frac{1}{110}+1\right)\)
A = (1 + 1 + 1 +...+ 1) + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
A = 10 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
A = \(10+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
A = \(10+\left(1-\frac{1}{11}\right)\)
A = \(10+\frac{10}{11}\)
A = \(\frac{120}{11}\)
58 + 92 -73:11
=58 + 92 -73/11
=150-73/11
=1577/11
25+72:(24+1)+72
=25+72:25+72
=25+2,88+72
=27,88+72
=99,88
\(=150-\frac{73}{11}=\frac{1577}{11}\)
\(=25+72:25+72=25+72:\left(25+1\right)=25+72:26=25+3=28\)
F = 7 + 72 + 73 + 74 + ..... + 7100
F= 7+(1+7)+73+(1+7)+...+799+(1+7)
F = 7x8+73x8+...+799x8
F= 8x(7+73+...+799)
mà 8 chia hết 8 => 8(7+73+...+799) chia hết 8
Vậy F chia hết cho 8
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)