202:5?
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\(202x\frac{3}{4}+202x\frac{5}{4}+202x8\)
\(=202x\left(\frac{3}{4}+\frac{5}{4}+8\right)\)
\(=202x\left(\frac{8}{4}+8\right)\)
\(=202x\left(2+8\right)\)
\(=202x10\)
\(=2020\)
\(=202*(\frac{3}{4}+ \frac{5}{4}+8) = 202*9 = 1818\)
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a) 407 + 24 + (-407) + (-84)
= [407 + (-407)] + 24 + (-84)
= 0 + 24 + (-84)
= 24 + (-84)
= (-60)
b) (-42) x 23 + 77 x (-42) - 200
= (-42) x (23 + 77) - 200
= (-42) x 100 - 200
= (-4200) - 200
= (-4400)
c) (-2021) + (499 + 2021) + 41
= (-2021) + 499 + 2021 + 41
= [(-2021) + 2021] + 499 + 41
= 0 + 499 + 41
= 499 + 41
= 540
d) (-202) - (34 - 202) - 66 + 50
= (-202) - 34 + 202 - 66 + 50
= [(-202) + 202] - 34 - 66 + 50
= 0 - 34 - 66 + 50
= (-34) - 66 + 50
= (-100) + 50
= (-50)
e) Bạn viết lại đề bài.
![](https://rs.olm.vn/images/avt/0.png?1311)
em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé
D = \(\dfrac{1}{7^2}\) - \(\dfrac{2}{7^3}\) + \(\dfrac{3}{7^4}\) - \(\dfrac{4}{7^5}\) +........+ \(\dfrac{201}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)
7 \(\times\) D = \(\dfrac{1}{7}\) - \(\dfrac{2}{7^2}\) + \(\dfrac{3}{7^3}\) - \(\dfrac{4}{7^4}\) + \(\dfrac{5}{7^5}\) -.......- \(\dfrac{202}{7^{202}}\)
7D +D = \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)
D = ( \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)) : 8
Đặt B = \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -........+\(\dfrac{1}{7^{201}}\).-\(\dfrac{1}{7^{202}}\)
7 \(\times\) B = 1 - \(\dfrac{1}{7}\)+\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) + \(\dfrac{1}{7^4}\) - \(\dfrac{1}{7^5}\) +.........- \(\dfrac{1}{7^{201}}\)
7B + B = 1 - \(\dfrac{1}{7^{202}}\)
B = ( 1 - \(\dfrac{1}{7^{202}}\)) : 8
D = [ ( 1 - \(\dfrac{1}{7^{202}}\)): 8 - \(\dfrac{202}{7^{203}}\)] : 8
D = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mỗi bài làm một nửa
bài 1: so sánh các lũy thừa
số trước =a số sau =b
a) 3^500 và 7^300
3^500=(3^5)^100=(27.9)^100
7^300=(7^3)^100=(49.7)^100
3^5<7^3
=>a<b
b) 8^5 và 3.4^7
8^5=2^3^5=2^15
3.4^7=3.2^14=2^15+2^14
a<b
bai 2: tìm chữ số tận cùng:
a)234^567
4^1=tận cùng =4
4^2=6
4^3=4
4^5=6
4^6=4
=>4^n tân cùng là 4 nếu n chẵn
=6 nếu n lẻ
567 lẻ=> đáp số =6.
a) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}\)
\(243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
b) \(8^5=\left(2^3\right)^5=2^{15}\)
\(3.4^7=3.\left(2^2\right)^7=3.2^{14}>2.2^{15}=2^{15}\)
\(\Rightarrow8^5< 3.4^7\)
\(202^{303}=\left(101.2\right)^{303}=101^{303}.2^{303}\)
\(=101^{202}.101^{101}.8^{101}=101^{202}.808^{101}\)
Mà
\(303^{202}=\left(3.101\right)^{202}=101^{202}.3^{202}=101^{202}.9^{101}\)
Dễ thấy \(101^{202}.808^{101}>101^{202}.9^{101}\)
Do đó \(202^{303}>303^{202}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
202:5 = 40,4
202:5=40,4
chuc ban hoc tot