so sánh A và B biết:
A=\(\left[0.8\cdot7+\left(0.8\right)^2\right]\cdot\left(1.25\cdot7-\frac{4}{5}\cdot1.25\right)-47.86\)
B=\(\frac{\left(1.09-0.29\right)\cdot\frac{5}{4}}{\left(18.9-16.65\right)\cdot\frac{8}{9}}\)
\(ChoA=\left[0,8.7+\left(0,8\right)^2\right]\left(1,25.7-\frac{4}{5}.1,25\right)+31,64\)
\(B=\frac{\left(1.09-0.29\right).\frac{5}{4}}{\left(18.9-16.65\right).\frac{8}{9}}\)
Hỏi A gấp mấy lần B?
\(A=\left[0,8.7\left(0,8\right)^2\right].\left(1,25.7-\frac{4}{5}.1,25\right)+31,64\)
\(=0,8.\left(7+0,8\right).1,25.\left(7-0,8\right)+31,64\)
\(=0,8.7,8.1,25.6,2+31,64\)
\(=6,24.7,75+31,64\)
\(=48,36+31,64=80\)
\(B=\frac{\left(1,09-0,29\right).\frac{5}{4}}{\left(18,9-16,65\right).\frac{8}{9}}=\frac{0,8.1,25}{2,25.\frac{8}{9}}=\frac{1}{2}\)
\(A:B=80:\frac{1}{2}=160\)
Vậy A gấp 160 lần B.
Tính tổng :
a) \(\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11}+...+\frac{1}{2013\cdot2015\cdot2017}\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{2017^2}\right)\)
c) \(\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2017}\right)\)
Tính nhanh :
\(A=\left(1-\frac{2}{6\cdot7}\right)\left(1-\frac{2}{7\cdot8}\right)\left(1-\frac{2}{8\cdot9}\right)\cdot\cdot\cdot\left(1-\frac{2}{51\cdot52}\right)\)
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot\cdot\cdot\left(1+\frac{1}{99\cdot101}\right)\)
đụ cha mi
mi trù ta thi rớt HK II mà ta giúp mày hả
mấy bài này cũng dễ ẹt nữa
đừng có mơ ta sẽ giúp mày
ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
\(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\) = ?
\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2.2.2.5.7.5.5.7.7.7}{2.5.7.7.2.5.7.7}=\frac{2.5}{1}=10\)
Ko biết có đúng ko
\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2^3.\left(5.5^2\right).\left(7.7^3\right)}{2^2.5^2.7^{2^2}}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
Tính giá trị biểu thức
\(E=\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
E = \(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2^2.2.5^2.5.7^4}{2^2.5^2.7^4}=2.5=10\)
\(E=\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
E=\(\frac{2^3.5.7.5^2.7^3}{2^2.5^27^4}=>E=\frac{2^3.5^3.7^4}{2^2.5^27^4}=>E=2.5=10\)10
\(lim\left(\frac{1}{2\cdot4}+\frac{1}{5\cdot7}+\frac{1}{8\cdot10}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+1\right)}\right)\)
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+2\right)}=\frac{n}{6n+4}\)
b.\(\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\cdot\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
Bài 1: Tính a) \(\left(\frac{11}{12}:\frac{44}{16}\right)\cdot\left(\frac{-1}{3}+\frac{1}{2}\right)\) b) \(\frac{\left(-5^2\right)\cdot\left(-5\right)^3\cdot16}{5^4\cdot\left(-2\right)^4}\) c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\)d) \(\left(\frac{-1}{2}+\frac{1}{3}\right)\cdot\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{4}{5}\)
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45