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A = 2o + 21 + 22 + ... + 22010
=> 2A = 21 + 22 + 23 + ... + 22010 + 22011
Mà A = 20 + 21 + 22 + ... + 22010
=> 2A - A = A = 1 + 22011
B = 1 + 3 + 32 + ... + 3100
=> 3B = 3 + 32 + 33 + ... + 3100 + 3101
Mà B = 1 + 3 + 32 + ... + 3100
=> 3B - B = 2B = 2 + 3101
=> B = ( 2 + 3101 ) : 2
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A=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
suy ra 4A=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)((n+3)-(n-1))
=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)
=n(n+1)(n+2)(n+3)
Đặt ak = k.(k+1).(k+2)
4a1 = 1.2.3.3-0.1.2.3
4a2 = 2.3.4.3-1.2.3.3
………….
4an-1 = (n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)
4an = n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)
Cộng từng vế n, ta được:
4(a1+a2+a3+………….+an) = n.(n+1).(n+2).(n+3)
4[1.2.3+2.3.4+3.4.5+………………..+n.(n+1).(n+2)] = n.(n+1).(n+2).(n+3)
=> A = \(\frac{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}{4}\)
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Gọi 1+2+22+....+2100 là A
Ta có:
A=1+2+22+....+2100
2A=2+22+23+...+2101
2A-A=(2+22+23+...+2101)-(1+2+22+23+...+2100)
A=2101-1
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a) 60-3(x-2)=51
3(x-2)=60-51
3(x-2)=9
x-2 = 9:3
x-2 = 3
x = 3+2
x 5
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Bài 1: \(35-3.\left|x\right|=5:\left(2^3-4\right)\)
\(35-3\left|x\right|=5:\left(8-4\right)\)
\(35-3.\left|x\right|=20\)
\(3.\left|x\right|=15\)
\(\left|x\right|=5\)
\(\Rightarrow x\in\left\{-5;5\right\}\)
Bài 2:
\(2017-\left(37+2017\right)+\left(-22+37\right)=2017-37-2017+\left(-22\right)+37\)
\(=\left(2017-2017\right)+\left(-37+37\right)+\left(-22\right)\)
\(=0+0+\left(-22\right)\)
\(=-22\)
trả lời nhanh giùm mình nha chỉ cần bài 1 thôi cx đc mấy bài kia mik biết làm rồi.Mai là mình thi òi nên mình cần cách làm để ôn thi
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Bài 1:
\(\frac{3}{5}+\frac{4}{15}=\frac{9}{15}+\frac{4}{15}=\frac{13}{15}\)
\(\frac{5}{6}:\frac{-7}{12}=\frac{5}{6}.\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
\(\frac{-21}{24}:\frac{-14}{8}=\frac{-21}{24}.\frac{-8}{14}=\frac{168}{336}=\frac{1}{2}\)
\(\frac{4}{5}:\frac{-8}{15}=\frac{4}{5}.\frac{-15}{8}=\frac{-60}{40}=\frac{-3}{2}\)
\(\frac{5}{12}-\frac{-7}{6}=\frac{5}{12}+\frac{7}{6}=\frac{5}{12}+\frac{14}{12}=\frac{19}{12}\)
\(\frac{-15}{16}.\frac{8}{25}=\frac{-120}{400}=\frac{-3}{10}\)
Bài 2 :
\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=\left(\frac{34}{5}-\frac{19}{5}\right)-\frac{5}{3}\)
\(=3-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(6\frac{5}{7}-\left(1\frac{2}{3}+2\frac{5}{7}\right)\)
\(=\frac{47}{7}-\left(\frac{5}{3}+\frac{19}{7}\right)\)
\(=\frac{47}{7}-\frac{5}{3}-\frac{19}{7}\)
\(=\left(\frac{47}{7}-\frac{19}{7}\right)-\frac{5}{3}\)
\(=4-\frac{5}{3}\)
\(=\frac{7}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\)
\(=\left(\frac{4}{19}+\frac{15}{19}\right).\frac{-3}{7}+\frac{5}{7}\)
\(=1.\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
A=\(2^0+2^1+2^2+2^3+...+\)\(2^{2010}\)
2A=\(2^1+2^2+2^3+...+2^{2011}\)
2A-A=\(2^1+2^2+2^3+...+2^{2011}\)-\(2^0-2^1-2^2-...-2^{2010}\)
A=\(2^{2011}-1\)
vậy A=\(2^{2011}-1\)
\(A=2^0+2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=1+2^1+2^2+2^3+2^4+...+2^{2010}\)
=> \(2A=2+2^2+2^3+2^4+2^5+...+2^{2011}\)
=> \(A=2^{2011}-1\)
Study well ! >_<