a,\(\dfrac{1996.2011-100}{996+2010.1996}\)
b,\(\dfrac{62.50+44.100}{27.38+146.19}\)
c,\(\dfrac{7116-14}{10290-35}\)
d,\(\dfrac{2929-101}{2.1919+404}\)
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\(A=\dfrac{4116-14}{10290-35}=\dfrac{14\times\left(294-1\right)}{35\times\left(294-1\right)}=\dfrac{14}{35}\)
\(B=\dfrac{2929-101}{2\times1919+404}=\dfrac{101\times\left(29-1\right)}{101\times\left(38+4\right)}=\dfrac{29-1}{38+4}=\dfrac{28}{32}=\dfrac{7}{8}\)
a) \(\dfrac{4116-14}{10290-35}=\dfrac{4102}{10255}=\dfrac{2051.2}{2051.5}=\dfrac{2}{5}\)
b) \(\dfrac{2929-101}{2.1919+404}=\dfrac{2929-101}{3838+404}=\dfrac{2828}{4242}=\dfrac{1414.2}{1414.3}=\dfrac{2}{3}\)
Rút gọn :
A= \(\dfrac{4116-14}{10290-35}\)= \(\dfrac{4102}{10255}=\dfrac{4102:2051}{10255:2051}=\dfrac{2}{5}\)
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
Các câu còn lại tương tự
\(a,\dfrac{2929-101}{2.1919+404}=\dfrac{29.101-101.1}{2.19.101+4.101}\)
\(=\dfrac{101\left(29-1\right)}{101\left(2.19+4\right)}\)
\(=\dfrac{101.29}{101.42}\)
\(=\dfrac{28}{42}=\dfrac{2}{3}\)
\(b,\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}\)
\(=\dfrac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}\)
\(=\dfrac{5.8.3.\left(-5\right).64}{5.3.9.\left(-2\right)^{14}}\)
\(=\dfrac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}\)
\(=\dfrac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\dfrac{5}{9.\left(-2\right)^5}\)
A=4116-14/10290-35
=[14 x ( 294 -1 )]/[35 x (294-1)]
= 14/35
B =2929-101/2.1919+404
= [101 x (291-1)] / [101 x 38 + 101 x 4)]
= [ 101x (29 -1)] / [ 101 x (38+4)
= (29-1) / (38+4)
= 28/42 = 2/3
A=4116-14/10290-35=14.294-14.1/35.294-35.1=14.(291-1)/35.(294-1)=14/35=14:7/35:7=2/5
B=2929-101/2.1919+404=101.29-101.1/2.19.101+4.101=101.(29-1)/101.(2.19+4)=28/42=28:14/42:14=2/3
b: \(=\dfrac{50\left(62+44\cdot2\right)}{38\left(27+73\right)}=\dfrac{50\cdot150}{38\cdot100}=\dfrac{75}{38}\)
c: \(=\dfrac{7112}{10255}=\dfrac{1016}{1465}\)