A=\(\frac{15^3+5\times15^2-5^3}{18^3+6\times18^2-6^3}\)
B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^{2015}}\)
C=\(\left(\frac{1}{2^2}-1\right)\times\left(\frac{1}{3^2}-1\right)\times.........\times\left(\frac{1}{100^2}-1\right)\)
tính nhanh :
\(A=\frac{15^3+5\times15^2-5^3}{18^3+6\times18^2-6^3}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2015}}\)
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\) \(D=\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(F=\left(100-1^2\right)\left(100-2^2\right)....\left(100-25^2\right)\)
Ta có: F= (100-12) (100-22)...(100-252)
=> F= (100-12)...(100-102)...(100-252)
=> F= (100-12)...0...(100-252)
=> F= 0
Vậy F= 0
1. Tính :
a.\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
b.\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{3}{7}\right)\times......\times\left(1-\frac{10}{7}\right)\)
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
= \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
= \(\frac{2}{3}+\frac{1}{11}\)
= \(\frac{25}{33}\)
tính giá trị biểu thức
a, A=\(\frac{-1}{2}-\left[\frac{-3}{5}\right]+\left[\frac{-1}{9}\right]+\frac{1}{27}+\frac{7}{18}+\frac{4}{35}-\left[-\frac{2}{7}\right]\)
b, B=\(\frac{1}{3}-\frac{3}{4}-\left[\frac{-3}{5}-\frac{1}{57}+\frac{1}{36}+\frac{-1}{15}\right]-\frac{2}{9}\)
c, C=\(\left[-\frac{7}{15}\right]\times\frac{5}{8}\times\left[\frac{30}{-7}\right]\times\left[-16\right]\times\left[\frac{-1}{1000}\right]\)
d, D=\(\frac{1}{2}\times\frac{-11}{19}-50\%\times\left[-\frac{1}{19}\right]+\frac{10}{19}\times\frac{1111}{2222}\)
tính giá trị biểu thức chứ còn cái gì nữa
a, \(A=\frac{22}{27}\)
b,\(B=\frac{1}{57}\)
C,\(C=\frac{1}{50}\)
d, \(D=0\)
BÀi 1: Thực hiện phép tính ( tính nhanh nếu có thể)
a.\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{71}-\left(-\frac{2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
b.\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
c.\(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{1}{3}-1\frac{1}{15}\right)\)
\(A=\left(1\frac{1}{6}\times\frac{6}{7}\times6:\frac{3}{5}\right):\left(4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{10}\right)\)
\(B=1\frac{13}{15}\times25\%\times3+\left(\frac{8}{15}-\frac{79}{60}\right):1\frac{23}{4}\)
\(C=\frac{123}{4567}\times\frac{1}{8}+\frac{123}{4567}\times\frac{1}{2}-\frac{123}{4567}\times\frac{13}{8}\)
\(D=\frac{10\frac{1}{3}\times\left(24\frac{1}{2}-15\frac{6}{7}\right)-\frac{12}{11}\times\left(\frac{10}{3}-1,75\right)}{\left(\frac{5}{9}-0,25\right)\times\frac{60}{11}+194\frac{8}{99}}\)
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{\left(3n-1\right)\times\left(3n+2\right)}=\frac{1}{6n+4}\)
b.\(\frac{5}{3\times7}+\frac{5}{7\times11}+\frac{5}{11\times15}+...+\frac{5}{\left(4n-1\right)\times\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)
Rút gọn:
a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\) b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)
a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\)
\(C=\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)^5}{\frac{2^2}{5^2}.\frac{\left(-5\right)^2}{12^2}}\)
\(C=\frac{\frac{-\left(2^3.3^2\right)}{3^3.2^4}}{\frac{2^2.5^2}{5^2.2^4.3^2}}\)
\(C=\frac{\frac{-1}{3.2}}{\frac{1}{2^2.3^2}}\)
\(C=\frac{\frac{-1}{6}}{\frac{1}{36}}\)
\(C=-6\)
b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^6+3^6}{-73}\)
\(D=\frac{3^6\left(2^6+2^3+1\right)}{-73}\)
\(D=\frac{3^6.73}{\left(-1\right).73}\)
\(D=-3^6=-729\)
bài 1 : tìm x biết
a, \(\frac{2}{3}\times\left(x-\frac{5}{6}\right)+\frac{1}{4}=\frac{22}{9}\)
b, \(\frac{2}{3}:\frac{x}{5}=\frac{10}{21}\)
c, \(\frac{7}{3}:\frac{x}{5}=\frac{14}{15}\)
d, \(1-\left(5\frac{4}{9}\times x-7\frac{7}{18}\right):15\frac{3}{4}=0\)
bài 2 : tính gtri bt
a,\(\frac{8}{7}+\frac{1}{5}\times\frac{10}{9}\)
b, \(\frac{3}{2}+\left(\frac{9}{2}+\frac{2}{9}\right)\times\left(\frac{4}{3}-\frac{5}{4}\right)\)
!_ove
a) x = 99/20
b) x = 7
c) x = 2
( chỉ lm đc đến đó thui nk )
Tính giá trị của biểu thức:
a) A= (153 + 5. 152 - 53) : ( 183 + 6. 182 - 63)
b) \(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\)
c) \(C=\frac{-1}{1.2}.\frac{-2^2}{2.3}.\frac{-3^2}{3.4}...\frac{-99^2}{99.100}.\frac{-100^2}{100.101}\)
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)