202 x 12,5 x 50% + 202 : 4/5 + 202 : 0,4 + 1
202 x 12,5 x 50 % + 202 : + 202 : 0,4 + 1
làm nhanh giúp mik nhé;>>>
=202*6,25+202*2,5+202*1,25+1
=2020+1
=2021
\(202x\frac{3}{4}+202x\frac{5}{4}+202x8\)
\(=202x\left(\frac{3}{4}+\frac{5}{4}+8\right)\)
\(=202x\left(\frac{8}{4}+8\right)\)
\(=202x\left(2+8\right)\)
\(=202x10\)
\(=2020\)
\(=202*(\frac{3}{4}+ \frac{5}{4}+8) = 202*9 = 1818\)
202 x \(\frac{3}{4}\) + 202 x\(\frac{5}{4}\) + 202 x 8
= 202 x ( \(\frac{3}{4}\)+\(\frac{5}{4}\)+ 8 )
= 202 x ( \(\frac{8}{4}\)+ 8 )
= 202 x ( 2 + 8 )
= 202 x 10
= 2020 .
a, so sánh : 202^303 và 303^202 ;3^21 và 2^31
b, tim n sao cho : 50<2^n<100 ; 50<7^n<2500
c, tim x biết : 2^x*7=224 ; (3*x+5)^2=289 ; x*(x ^2)^3=x^5 ; 3^2*x+1 *11=2673
Nhiều quá
a)202303=(2023)101=8242408101
303^202=(302^2)^101=91204^101
Vì 824208^101>91204^101=>202^303>303^202
Tính bằng cách thuận tiện:
a) 2019 x 2 x 5
b) 5 x 8 x 9 x 2 x 125
c) 4 x 2019 x 25
d) 50 x 125 x 2 x 8 x 202
a, 2019 x (2 x 5)
= 2019 x 10
= 20 190
b, (5 x 2) x (8 x 125) x 9
= 10 x 1000 x 9
= 90 000
c, (4 x 25) x 2019
= 100 x 2019
= 201 900
d, (50 x 2) x (125 x 8) x 202
= 100 x 1000 x 202
= 20 200 000
a) 2019 x 2 x 5
= (2 x 5) x 2019
= 10 x 2019
= 20190b) 5 x 8 x 9 x 2 x 125
= (5 x 2) x (125 x 8) x 9
= 10 x 1000 x 9
= 10000 x 9
= 90000
c) 4 x 2019 x 25
= (4 x 25) x 2019
= 100 x 2019
= 201900
d) 50 x 125 x 2 x 8 x 202
= (50 x 2) x (125 x 8) x 202
= 100 x 1000 x 202
= 100000 x 202
= 20200000
Tính bằng cách thuận tiện:
50 x 125 x 2 x 8 x 202
50 x 125 x 2 x 8 x 202
= (50 x 2) x (125 x 8) x 202
= 100 x 1000 x 202
= 100000 x 202
= 20200000
=(50x2)x(125x8)x202
=100x1000x202=100 000x202=202 00000
x - 1+ x - 3 + x - 5 + x -7 + … + x -201= 202
407+24+(-407)+(-84) (-42)nhân 23+77nhân(-42) - 200. (-2021)+(499 + 2021)+41. (-202)-(34-202) - 66 + 50. 900 : (300 : (340 -(5 nhân 8 ậm - 4âm2 nhân 5)
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.
\(\dfrac{1}{5+1}\)+\(\dfrac{2}{5^2+1}\)+\(\dfrac{3}{5^3+1}\)+....+\(\dfrac{202}{5^{202}+1}\) < \(\dfrac{1}{4}\)
Chứng minh giúp em ạ!!!
\(\dfrac{1}{5+1}\)
cho D=1/7^2-2/7^3+3/7^4-4/7^5+.....+201/7^202-202/7^203. Hãy so sánh D với 1/64.
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 = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)