Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)
\(\Leftrightarrow A=365\div365=1\)
Vậy \(A=1\)
b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)
\(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)
Vậy \(B=0\)
c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)
\(\Leftrightarrow C=2\)
Vậy \(C=2\)
d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)
- Ta có: \(D=1152-374-1152-65+374\)
\(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)
\(\Leftrightarrow D=-65\)
Vậy \(D=-65\)
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
Câu 1 dễ mà :
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8
= 1.2.3...8.( 9 - 1 - 8 )
= 1.2.3...8.0
= 0
a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)
Vậy A = 3
b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)
\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)
Vậy B = 8
c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)
\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)
Vậy C = 6
d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)
Vậy D = 2
\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}\)
\(N=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
N=1/2x(1/3-1/5+1/5-1/7+....+1/99-1/101)
N=1/2x(1/3-1/101)
N=1/2x98/101
N=49/101
bạn phải cho ra 2 số cuối thì mới làm đc nha có 1 s
ố cuối ko làm đc đâu
A= 1-1/2 + 1-1/3 + 1/2-1/5 + 1/3-1/8+ 1/5-1/13+1/8- 1/21 +....+ 1/610- 1/1597
A= 1/610
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
c) \(C=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100\)
\(\Rightarrow C=\frac{98.99.100}{3}=323400\)
d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)