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(99-97)+(95-93)+.....................+(7-5)+(3-1)
=>(99-1):2+1+50
<=>50:2=25
mà 25x2=50
Bài 1:
a) 81 + 243 + 319 = 81 + 319 + 243
= 400 + 243 = 643
b) 168 + 1999 + 132 = 168 + 132 + 1999
= 300 + 1999 = 2299
c) 5.25.74.20.4 = (5.20).(25.4).74
= 100.100.74 = 740 000
d) 32.74 + 32.53 = 32.( 74 + 53 )
= 32.127 = 4064
e) 26 + 27 + 28 + 29 + ... + 34 + 35 = (26 + 34) + (27+33) + (28 + 32) + (29 + 31) + 30 + 35
= 60 x 4 + 30 + 35
= 240 + 30 + 35 = 270 + 35 = 305
g) 1 + 3 + 5 + 7 + ... + 99 + 101 = (1 + 99) + (2 + 98) + ... + (48 + 52) + (49 + 51) + 101
= 100 x 49 + 101
= 490 + 101 = 591
Bài 2:
a) 23 - ( 34 - x ) = 23 b) x - ( 74 + 126 ) = 100
34 - x = 23 - 23 74 + 126 = x - 100
34 - x = 0 200 = x - 100
x = 34 - 0 x = 200 - 100
x = 34 x = 100
c) 2436 : x = 12 d) 317 + ( 146 - x ) = 417
x = 2436 : 12 146 - x = 417 - 317
x = 23 146 -x = 100
x = 146 - 100
x = 46
Mình không có thời gian để làm nốt câu e, mong bạn thông cảm!
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
E = 1.2 + 2.3 + 3.4 + … + 99.100
=> 3E = 3.( 1.2 + 2.3 + 3.4 + … + 99.100)
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
=> 3E = 1.2.3 + 2.3.(4 – 1) + 3.4.(5 – 2) + … + 99.100.(1011 –98 )
=> 3E = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + … + 99.100.101– 98.99.100
=> 3E = 99.100.101
=> E = \(\frac{99.100.101}{3}=333300\)
=> 3E = 1.2.3 + 2.3.(4 – 1) + 3.4.(5 – 2) + … + .99.100.(101-98)
nha , sửa hộ , ghi thừa 1 số 1 dòng số 5 có số 1011 thì sửa thành 101 nha
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha