\(\frac{\left(-5\right)^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\cdot\left(-100\right)^0}\)
TT
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a) frac{2929-101}{2cdot1919+404} b)frac{-1997cdot1996+1}{left(-1995right)cdot left(-1997right)+1996} c)frac{left(-5right)^3cdot40cdot4^3}{135cdotleft(-2right)^{14}cdotleft(-100right)^0} d)frac{3cdot7cdot13cdot37cdot39-10101}{505050-70707}
Đọc tiếp
a) \(\frac{2929-101}{2\cdot1919+404}\) b)\(\frac{-1997\cdot1996+1}{\left(-1995\right)\cdot \left(-1997\right)+1996}\)
c)\(\frac{\left(-5\right)^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\cdot\left(-100\right)^0}\) d)\(\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)
Tính giá trị của biểu thức:a,(32)2-(-23)2-(-52)3b,left|frac{-1}{2}right|^2cdotleft(-32right)-left(-8right)+left|frac{1}{2}right|^3c,2^3+3cdotleft(frac{-5}{86}right)^0cdotleft(frac{1}{2}right)^2cdot4+left[left(-2right)^2:frac{1}{2}right]:8d,left|frac{5}{7}cdotleft(-14right)right|-left(frac{2}{3}right)^2cdotleft(-18right)+6^2cdotfrac{-1}{18}
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Tính giá trị của biểu thức:
a,(32)2-(-23)2-(-52)3
b,\(\left|\frac{-1}{2}\right|^2\cdot\left(-32\right)-\left(-8\right)+\left|\frac{1}{2}\right|^3\)
c,\(2^3+3\cdot\left(\frac{-5}{86}\right)^0\cdot\left(\frac{1}{2}\right)^2\cdot4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)
d,\(\left|\frac{5}{7}\cdot\left(-14\right)\right|-\left(\frac{2}{3}\right)^2\cdot\left(-18\right)+6^2\cdot\frac{-1}{18}\)
\(C=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot....\cdot\left(1-\frac{2}{99\cdot100}\right)\)
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)
Chứng minh rằng:
a)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)
b)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot\cdot\cdot2n}=\frac{1}{2^n}\)Với \(n\inℕ^∗\)
G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
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\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
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1/Sleft(1+dfrac{1}{2}right)cdotleft(1+dfrac{1}{3}right)cdotleft(1+dfrac{1}{4}right)cdot...cdotleft(1+dfrac{1}{100}right)
2/Bleft(1-dfrac{1}{2}right)cdotleft(1-dfrac{1}{3}right)cdotleft(1-dfrac{1}{4}right)cdot...cdotleft(1-dfrac{1}{2007}right)
3/Cdfrac{2^2}{1cdot3}cdotdfrac{3^2}{2cdot4}cdotdfrac{4^2}{3cdot5}cdot...cdotdfrac{100^2}{99cdot101}
Đọc tiếp
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
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Tính
A=\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot....\cdot\left(1-\frac{1}{1326}\right)\)
B=\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\frac{1}{99\cdot101}\right)\)
a)left(2^2+3right)cdotleft(x-5right)+145^2124:2^2 b) 3^2cdotleft(x+1right)-32^3+7^2cdot2:14c) 2^2cdot3cdotleft(x+5right)-6left(2^3+2^2right)cdot2^2d) left(2^2+1right)cdotleft(x+14right)5^2cdot4+left(2^5+3^2+7^2right):2e) left(2^2-1right)cdotleft(x-1right)2^2+left(6^2+2^6right):left(5^2cdot2right)g) left(3^2-2right)cdotleft(x-12right)+355^2+279:3^2nhìn thì cảm thấy khó nhưng lại rất dễ đó
Đọc tiếp
a)\(\left(2^2+3\right)\cdot\left(x-5\right)+14=5^2124:2^2\)
b) \(3^2\cdot\left(x+1\right)-3=2^3+7^2\cdot2:14\)
c) \(2^2\cdot3\cdot\left(x+5\right)-6=\left(2^3+2^2\right)\cdot2^2\)
d) \(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
e) \(\left(2^2-1\right)\cdot\left(x-1\right)=2^2+\left(6^2+2^6\right):\left(5^2\cdot2\right)\)
g) \(\left(3^2-2\right)\cdot\left(x-12\right)+35=5^2+279:3^2\)
nhìn thì cảm thấy khó nhưng lại rất dễ đó
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