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Ta có : A = 1 + 6 + 6^2 + .... + 6^9 .
= 1 + 6 . ( 1 + 6 + ..... + 6^8 ) .
Do đó A chia cho 6 dư 1
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A< 1-\frac{1}{10}=\frac{9}{10}\)
\(=>A>\frac{65}{132}\)
Answer:
Bài 1:
\(\left(135-35\right).\left(-37\right)+37.\left(-42-58\right)\)
\(=100.\left(-37\right)+37.\left(-100\right)\)
\(=\left(-3700\right)+\left(-3700\right)\)
\(=-7400\)
\(-65.\left(87-17\right)-87.\left(17-65\right)\)
\(=\left(-65\right).87-\left(-65\right).17-87.17+87.65\)
\(=-\left(-65\right).17-87.17\)
\(=17.\left(65-87\right)\)
\(=-374\)
\([3.\left(-2\right)-\left(-8\right)].\left(-7\right)-\left(-2\right).\left(-5\right)\)
\(=[\left(-6\right)+8].\left(-7\right)-\left(-2\right).\left(-5\right)\)
\(=2.\left(-7\right)-10\)
\(=-24\)
Bài 2:
\(\left(-55\right).\left(-25\right).\left(-8\right)\)
\(=\left(-55\right).[\left(-25\right).\left(-8\right)]\)
\(=\left(-55\right).200\)
\(=-11000\)
\(\left(-1\right).\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right).10\)
\(=[\left(-1\right).10].[\left(-2\right).\left(-5\right)].[\left(-3\right).\left(-4\right)]\)
\(=\left(-10\right).10.12\)
\(=-1200\)
Chị gái xinh đẹp à. Câu hỏi của chị khó quá ko ai trả lời. Thôi thì.......k cho mem đi😉
có 2 truongf hợp
* x+7=X-9 suy rA ko tìm đc x
* x+7=-(x-9) suy ra x=1
- Giải:
a) 29 + ( 132 + 868 ) + ( 237 + 763 )
= 29 + ( 1000 + 1000 )
= 29 + 2000 = 2029
b) 652 + 327 + 148 + 15 + 73
= ( 652 + 148 ) + ( 327 + 73 ) + 15
= ( 800 + 400 ) + 15 = 1215
-------Phần còn lại nhờ mấy bn khác lm hộ nhé-------
a) 29 + 132 + 237 + 868 + 763
= 29 + (132 + 868) + (237 + 763)
= 29 + 1000 + 1000
= 2029
b) 652 + 327 + 148 + 15 + 73
= (652 + 148) + (327 + 73) + 15
= 800 + 400 + 15
= 1215
c) 35 . 24 + 35 . 38 + 65 . 75 + 45 . 65
= 35(24 + 38) + 65(75 + 45)
= 35 . 62 + 65 . 120
= 5 . 7 . 2 . 31 + 65 . 12 . 10
= 10 . 7 . 31 + 65 . 12 .10
= 10(7 . 31 + 65 . 12)
= 10(217 + 780)
= 10 . 997
= 9970
d) 3 . 25 + 4 . 37 . 6 + 2 . 38 . 12
= 3 . 25 + 4 . 37 . 2 . 3 + 2 . 38 . 12
= 3 . 25 + 12 . 37 . 2 + 2 . 38 . 12
= 3 . 25 + 12 . 2(37 + 38)
= 3 . 25 + 12 . 2 . 75
= 3 . 25 + 12 . 2 . 3 . 25
= 3 . 25(1 + 12 . 2)
= 3 . 25 . 25
= 1875
a, (3x +)^3 =4^3 b, (2x +1 )^3 =5^3
<=>3x +1 =4 <=>2x +1 =5
<=>x=1 <=>x=2
\(\left(\dfrac{1}{4}+35\%+0,65+75\%\right)\cdot\left(1\dfrac{6}{9}+4\dfrac{8}{24}\right)\\ =\left(\dfrac{1}{4}+\dfrac{7}{20}+\dfrac{13}{20}+\dfrac{3}{4}\right)\cdot\left(\dfrac{5}{3}+\dfrac{13}{3}\right)\\ =\left[\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{7}{20}+\dfrac{13}{20}\right)\right]\cdot\left(\dfrac{18}{3}\right)\\ =\left(\dfrac{4}{4}+\dfrac{20}{20}\right)\cdot6\\ =\left(1+1\right)\cdot6\\ =2\cdot6\\ =12\)
\(\left(\dfrac{1}{4}+35\%+0,65+75\%\right)\left(1\dfrac{6}{9}+4\dfrac{8}{24}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{7}{20}+\dfrac{13}{20}+\dfrac{3}{4}\right)\left(\dfrac{5}{3}+\dfrac{13}{3}\right)\)
\(=\left\{\left[\dfrac{1}{4}+\dfrac{3}{4}\right]+\left[\dfrac{7}{20}+\dfrac{13}{20}\right]\right\}\left(\dfrac{5+13}{3}\right)\)
=\(\left(1+1\right).6=2.6=12\)