a). -8 .{[62 : (-12 : 4)] + 3} =
b). (\(2\frac{1}{3}\)- \(\frac{1}{9}\)) : (-\(2\frac{1}{6}\) + 0,25 ) + \(1\frac{11}{69}\)=
tính
a, \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b , \(\left(\frac{0,4-\frac{8}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
c, A
= \(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+...+\frac{1}{20}.\left(1+2+3+....+20\right)\right).155\)
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
\(a,1\frac{13}{15}.0,75-\left(\frac{8}{15}+0,25\right).\frac{24}{47}\)
\(b,5:\left(4\frac{3}{4}-1\frac{25}{28}\right)-1\frac{3}{8}:\left(\frac{3}{8}+\frac{9}{20}\right)\)
\(c,6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(d,\left(\frac{3}{5}+0,415-\frac{3}{200}\right).2\frac{2}{3}.0,25\)
\(e,\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(g,1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right);\frac{7}{3}\)
a )\(3^4.\left(-4\right)^2\)
b )\(|-1\frac{3}{5}|+(6\frac{5}{11}-4\frac{5}{11}).\frac{1}{5}\)
c )\(\left(-2\right)^2.(\frac{3}{4}-0,25)\)
d ) \(\frac{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}{\frac{4}{12}+\frac{9}{12}+\frac{18}{12}}\)
Nhanh ! CẦN GẤP!
a) \(3^4.\left(-4\right)^2=3^4.4^2=3^4.2^4=\left(3.2\right)^4=6^4\)
b) \(=1\frac{3}{5}+\left(6+\frac{5}{11}-4-\frac{5}{11}\right).\frac{1}{5}=1+\frac{3}{5}+2.\frac{1}{5}=2\)
c)\(=4.\left(0,75-0,25\right)=4.0,5=2\)
d) \(=\frac{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}{\frac{1.4}{3.4}+\frac{3.3}{4.3}+\frac{3.6}{2.6}}=\frac{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}=1\)
a. \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)
b. \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)
c. \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)
d. \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)
e. \(-1,5\cdot\left(1+\frac{2}{3}\right)\)
f. \(1^2+2^2+3^2+...+100^2\)
a, \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)
\(=\left(\frac{1}{4}+\frac{5}{12}\right)-\left(\frac{1}{13}+\frac{7}{8}\right)\)
\(=\frac{2}{3}-\frac{99}{104}\)
\(=-\frac{89}{312}\)
b, \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)
\(=\left(11\frac{3}{13}+5\frac{3}{13}\right)-2\frac{4}{7}\)
\(=\frac{214}{13}-\frac{18}{7}\)
\(=\frac{1264}{91}\)
c, \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)
\(=6\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)
\(=\left(6\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)
\(=2+3\frac{7}{11}\)
\(=5\frac{7}{11}\)
\(=\frac{62}{11}\)
d, \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)
\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\cdot0\)
\(=0\)
e, \(-1,5\cdot\left(1+\frac{2}{3}\right)\)
\(=-\frac{3}{2}\cdot\frac{5}{3}\)
\(=-\frac{5}{2}\)
f, Đặt \(A=1^2+2^2+3^2+...+100^2\)
\(=1+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)
\(=1+2\cdot3-2+3\cdot4-3+...+100\cdot101-100\)
\(=\left(2\cdot3+3\cdot4+...+100\cdot101\right)-\left(1+2+3+...+100\right)\)
Đặt B = 2 . 3 + 3 . 4 + ... + 100 . 101
3B = 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 100 . 101 . ( 102 - 99 )
3B = 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 100 . 101 . 102 - 99 . 100 . 101
3B = 100 . 101 . 102
B = \(\frac{100\cdot101\cdot102}{3}\)
B = 343400
Thay B vào A. Ta được :
\(A=343400-\left(1+2+3+...+100\right)\)
Thay C = 1 + 2 + 3 + ... + 100
Dãy số 1; 2; 3; ...; 100 có số số hạng là:
( 100 - 1 ) : 1 + 1 = 100 ( số hạng )
Tổng của dãy số đó là :
( 100 + 1 ) . 100 : 2 = 5050
=> C = 5050
Thay C vào A. Ta được :
\(A=343400-5050\)
\(A=338350\)
Vậy A = 338350
\( \left(2\frac{1}{3}-\frac{1}{9}\right):\left(-2\frac{1}{6}+0,25\right)+1\frac{11}{69}\)
giúp tôi giải bài này với :\(\left(2\frac{1}{3}-\frac{1}{9}\right):\left(-2\frac{1}{6}+0,25\right)+1\frac{11}{69}\)
BÀI ÔN SỐ HỌC
1. Thực hiện mỗi phép tính sau bằng 2 cách:
a)\(3\frac{4}{9}+5\frac{1}{6}\)
Ví dụ :
Cách 1: \(3\frac{4}{9}+5\frac{1}{6}=\frac{31}{9}+\frac{31}{6}=\frac{62}{18}+\frac{93}{18}=\frac{155}{18}=8\frac{11}{18}\)
Cách 2:\(3\frac{4}{9}+5\frac{1}{6}=3\frac{8}{18}+5\frac{3}{18}=8\frac{11}{18}\)
b)\(8\frac{1}{14}-6\frac{3}{7}\)
c)\(7-3\frac{6}{7}\)
2. Áp dụng tính chất phép tính & qui tắc dấu ngoặc để tính giá trị các biểu thức sau:
a)A =\(11\frac{3}{13}-\left(3\frac{4}{7}+6\frac{3}{13}\right)\)
b)B =\(\left(7\frac{4}{9}+3\frac{8}{13}\right)-5\frac{4}{3}\)
c)C =\(\frac{-2}{7}.\frac{5}{11}+\frac{-2}{7}.\frac{6}{11}+5\frac{4}{7}\)
d)D =\(0,7.1\frac{3}{5}.30.0,375.\frac{4}{7}\)
3.Tìm các số nghịch đảo của các số sau : \(\frac{4}{7};6\frac{3}{8};\frac{-3}{17};0.37\)
4.Tìm x, biết :
0,5-\(\frac{2}{3}x=\frac{5}{12}\)
Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều
a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)
b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)
c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
Bài 1 thực hiện phép tính
a)\(\frac{45}{19}-\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^{-1}\right)^{-1}.\)
b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}.\)
Bài 2. tìm x, biết:
a) 2(x-1) - 3(2x+2) - 4(2x+3) =16
b) \(3\frac{1}{2}:\left|2x-1\right|=\frac{21}{22}\)
c) |x2+|x-1|| = x2+2
Bài 3. Chứng minh rằng số có dạng abcabc luôn chia hết cho 11
Bài 4.tính:
a) A = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
b) B =\(4.\left(-\frac{1}{2}\right)^2-2.\left(-\frac{1}{2}\right)^2+3.\left(-\frac{1}{2}\right)+1\)
c) C =\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\)
d) D =\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)