13256 + 36565557 +23974 + 3 - 35 = ...............................
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a) \(S=1^5+3^5+....+75^5+99^5\)
\(\left(2a+1\right)^5-\left(2a+1\right)=2a\left(2a+1\right)\left(2a+2\right)\left[\left(2a+1\right)^2+1\right]\)
\(\left(2a+1\right)^5-\left(2a+1\right)=4a\left(2a+1\right)\left(a+1\right)\left[\left(2a+1\right)^2+1\right]⋮4\)
\(S=\left(1^5-1\right)+\left(3^5-3\right)+....+\left(75^5-75\right)+\left(99^5-99\right)+\left(1+3+5+...+75+99\right)\)
\(\Leftrightarrow\begin{matrix}1^5-1⋮4\\3^5-3⋮4\\5^5-5⋮4\\...........\\75^5-5⋮4\\99^5-99⋮4\end{matrix}\)
\(S_1=1+3+5+7+...+75+99=\frac{\left(1+75\right)\left[\frac{75-1}{2}+1\right]}{2}+99=38.38+96+3\)
\(\Rightarrow S_1:4\) dư 3
\(\Leftrightarrow S\) chia 4 dư 3
1. Mẫu số chung là 3x4x5=60
1/3 = 1x4x5/60 = 20/60
2/4 = 2x3x5 = 30/60
4/5 = 4x3x4 = 48/60
2. Tính nhanh 35x26+35x27+70/(35+35+...+35)
= 35x(26+27)+35x2/35x100
=35x53+1/50
=(35x53x50+1)/50
= 92751/50
Sửa đề
\(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{-9}{35}\)
\(x+2=35\)
\(x=35-2\)
\(x=33\)
\(-17x=-17-\left(-34\right)\)
\(-17x=-17\)
\(x=-1\)
Bạn coi lại đề: $\frac{3}{69}$ không nằm trong các số hạng có quy luật.
13256 + 36565557 +23974 + 3 - 35 = 36602755
13256+36565557+23974+3-35
=(13256+23974) gfh