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12,12 x 7,5 + 12,12 : 2 + 12,12 + 12,12
= 12,12 x 7,5 + 12,12 x 0,5 + 12,12 x 1 + 12,12 x 1
= 12,12 x (7,5 + 0,5 + 1 + 1)
= 12,12 x 10
= 121,2
Bài 9
a; A = (1 + \(\dfrac{1}{3}\))\(\times\)(1 + \(\dfrac{1}{8}\))\(\times\)(1 + \(\dfrac{1}{15}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{9999}\))
A = \(\dfrac{3+1}{3}\)\(\times\)\(\dfrac{8+1}{8}\)\(\times\)\(\dfrac{15+1}{15}\)\(\times\)...\(\times\)\(\dfrac{9999+1}{9999}\)
A = \(\dfrac{4}{3}\)\(\times\)\(\dfrac{9}{8}\)\(\times\)\(\dfrac{16}{15}\)\(\times\)...\(\times\)\(\dfrac{10000}{9999}\)
A = \(\dfrac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)
A = \(\dfrac{2\times100}{1\times101}\)
A = \(\dfrac{200}{101}\)
b; B = (1 - \(\dfrac{1}{4}\))\(\times\) (1 - \(\dfrac{1}{9}\)) \(\times\) (1 - \(\dfrac{1}{16}\))\(\times\)...\(\times\)(1 - \(\dfrac{1}{10000}\))
B = \(\dfrac{4-1}{4}\) \(\times\) \(\dfrac{9-1}{9}\) \(\times\) \(\dfrac{16-1}{16}\) \(\times\) \(\dfrac{10000-1}{10000}\)
B = \(\dfrac{3}{4}\)\(\times\) \(\dfrac{8}{9}\)\(\times\)\(\dfrac{15}{16}\)\(\times\)...\(\times\) \(\dfrac{9999}{10000}\)
B = \(\dfrac{1\times3}{2\times2}\)\(\times\)\(\dfrac{2\times4}{3\times3}\)\(\times\)\(\dfrac{3\times5}{4\times4}\)\(\times\)...\(\times\)\(\dfrac{99\times101}{100\times100}\)
B = \(\dfrac{1\times101}{2\times100}\)
B = \(\dfrac{101}{200}\)
Câu 1:
0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3
= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059
= 9 x 43,6 + 9 x 85,8 + 6 x 105,9
= 3 x 130,8 + 3 x 257,4 + 3 x 211,8
= 3 x ( 130,8 + 257,4 + 211,8 )
= 3 x 600
= 1800
Câu 2:
3/4 x X + 1/2 x X - 15 = 35
X x ( 3/4 + 1/2 ) - 15 = 35
X x ( 3/4 + 1/2 ) = 50
X x 5/4 = 50
X = 40
VẬy X = 40
Bài 1:
a.
$545,26+117,3=662,56$
b.
$400,56-184,48=216,08$
c.
$4,21\times 3,2=13,472$
d.
$28,5:2,5=11,4$
Bài 2:
a. 2km 21 m = 2,021 km
b. 1020 kg = 1 tấn 20 kg
c. 22 dam2 10 m2 = 2210 m2
d. 90 giây = 1,5 phút
S_AMD = \(\frac{1}{3}\) ABD (Chung chiều cao từ D, đáy AM = \(\frac{1}{3}\) AB)
Tương tự S_ BCP =\(\frac{1}{3}\) BCD. Mà S_(ABD + BCP) = S_ABCD => S_(AMD + BCP) = \(\frac{1}{3}\) ABCD
Nên S_MBPD = \(\frac{2}{3}\) ABCD => S_MPQ = \(\frac{1}{2}\) MPD (chung đường cao từ M đáy DP mà DQ = \(\frac{1}{2}\) DP)
Tương tự MNP = \(\frac{1}{2}\) MBP. Mà MBP + MPD = S_MBPD => S_(MPQ+MNP) = \(\frac{1}{2}\) S_MBPD
Hay S_MNPQ = \(\frac{1}{2}\) MBPD Mà MBPD = \(\frac{2}{3}\) ABCD
=> S_MNPQ = \(\frac{2}{3}.\frac{1}{2}\) ABCD = \(\frac{1}{3}\) ABCD
Vậy S_MNPQ = 480 : 3 = 160 (cm2)