29×73+27×29+3×15
Ai giải hộ mình
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ta có: A = 529 + 528 + 527 + ...+ 53 + 52 + 5 + 1
=> 5A = 530 + 529 + 528 + ...+ 54 + 53 + 52 + 5
=> 5A - A = 530 - 1
\(A=\frac{5^{30}-1}{4}\)
\(A=5^{29}+5^{28}+5^{27}+...+5^3+5^2+5+1\)
=> \(5A=5^{30}+5^{29}+5^{28}+....+5^4+5^3+5^2+5\)
=> \(5A-A=5^{30}-1\)
=> \(4A=5^{30}-1\) => \(A=\frac{5^{30}-1}{4}\)
a, 45 x 46 + 45 x 45 - 30 - 15= 45x( 46+45) -30 -15= 45 x 91 - 30 - 15= 4095 - 30 - 15= 4050
b, 891 x 29 + 29 x 8+ 29= 29 x ( 891 + 8+ 1)= 29 x 900= 26100
c, 73 x ( 84+17) + 27 x (74+27)= 73 x 101 + 27 x 101= 101 x ( 73+27)= 101 x 100= 10100
d, chịu
d, đề bài= 13 x ( 99 + 2) + 13 x 1001+ 11 x 1300= 13 x 101 + 13 x 1001+ 13 x 1100= 13 x ( 101 + 1001 + 1100) = 13 x 2202= 28626
\(\dfrac{4}{7}v\text{à }\dfrac{16}{63}\\ \dfrac{4}{7}=\dfrac{4\cdot9}{7\cdot9}=\dfrac{36}{63}\\ \dfrac{36}{63}>\dfrac{16}{63}\\ \Rightarrow\dfrac{4}{7}>\dfrac{16}{36}\)
\(\dfrac{4}{17}\) và \(\dfrac{16}{63}\)
\(\dfrac{4}{63}>\dfrac{16}{63}\)
\(=>\dfrac{4}{17}>\dfrac{16}{63}\)
\(\dfrac{5}{29}\) và \(\dfrac{7}{33}\)
\(\dfrac{5}{33}< \dfrac{7}{33}\)
\(=>\dfrac{5}{29}< \dfrac{7}{33}\)
\(\dfrac{44}{57}\) và \(\dfrac{89}{99}\)
\(\dfrac{44}{99}< \dfrac{89}{99}\)
\(=>\dfrac{44}{57}< \dfrac{89}{99}\)
\(\dfrac{19}{53}\) và \(\dfrac{30}{73}\)
\(\dfrac{19}{73}>\dfrac{30}{73}\)
\(=>\dfrac{19}{53}>\dfrac{30}{73}\)
Lời giải:
$22+23-25+27-29+31-33$
$=22+(23-25)+(27-29)+(31-33)$
$=22+(-2)+(-2)+(-2)=22+(-2).3=22-6=16$
a, \(\dfrac{90}{37}-\dfrac{38}{25}-\dfrac{8}{25}-\dfrac{4}{25}\)
= \(\dfrac{90}{37}\) - \(\dfrac{38+8+4}{25}\)
= \(\dfrac{90}{37}\) - 2
= \(\dfrac{16}{37}\)
\(\dfrac{24}{29}\) + \(\dfrac{32}{41}\) + \(\dfrac{34}{29}\) + \(\dfrac{50}{41}\)
=(\(\dfrac{24}{29}\) + \(\dfrac{34}{29}\)) + (\(\dfrac{32}{41}\) + \(\dfrac{50}{41}\))
= \(\dfrac{58}{29}\) + \(\dfrac{82}{41}\)
= 2 + 2
= 4
\(29\times73+27\times29+3\times15\)
\(=29\times\left(73+27\right)+3\times15\)
\(=29\times100+45\)
\(=2900+45\)
\(=2945\)
29x73+27x29+3x15
=29(73+27)+45
=29x100+45
=2900+45
=2945
Tk mk nha, thanks.