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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, 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
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 \(D=-65\)
\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)
a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)
\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)
Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015
\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)
<=>S=-1007+2015
<=> S=1008
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)
\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)
\(A=\frac{154}{11}+\frac{8}{5}\)
\(\Rightarrow A=\frac{78}{5}\)
\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)
\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)
\(B=\frac{27}{9}+\frac{-72}{13}\)
\(B=3+\frac{-72}{13}\)
\(\Rightarrow B=\frac{-33}{13}\)
P/s: Hoq chắc :v
a ) Ta thấy :
2^4 = 16
4^2 = 16
16 - 16 = 0
Số nào nhân với 0 cũng bằng 0 nên giá trị biểu thức trên là 0
b ) ( 7^2015 + 7^2014 ) : 7^2013
= 7^2015 : 7^2013 + 7^2014 : 7^2013
= 7^2 + 7
= 49 + 7
= 56
c ) ( 3 . 4 . 2^16 ) ^ 2 / 11 . 2^13 . 4^11 - 16^9
Tính phần mẫu trước .
11 . 2^13 . 4^11 - 16^9 = 11 . 2^13 . ( 2^2 ) ^11 - (2^4)^9 = 11 . 2^13 . 2^22 - 2^36 = 11. 2^35 - 2^36 = 11 . 2^35 - 2^35 . 2 = ( 11 - 2 ) . 2^35 = 9 . 2^35
Phần tử :
( 3 . 4 . 2^16 ) ^ 2 = 3^2 . ( 2^2 ) ^ 2 . ( 2^16 ) ^ 2 = 3 ^ 2 . 2^4 . 2^32 = 9 . 2^36
Vì các thừa số của mẫu và tử đều giống nhau nên có kết quả là 1 .
o0o I love you o0o nói đúng