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\(2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5\left[300-10^2\right]\)
\(=2011+5\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
học tốt
- 2011+5[300-(17-7)2]
- =2011+5[300-100]
- =2011+5+200
- =2011+205
=2216
\(a)47-\left[\left(45.2^4-5^2.12\right):14\right]\)
\(=47-\left[\left(45.16-25.12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30\)
\(=17\)
\(b)50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right)\right]:2+34\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40\)
\(=10\)
\(c)10^2-\left[60:\left(5^6:5^4-3,5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3,5\right)\right]\)
\(=10^2-\left[60:\left(25-3,5\right)\right]\)
\(=10^2-\left[60:21,5\right]\)
\(=100-\dfrac{120}{43}\)
\(=\dfrac{4180}{43}\)
\(d)50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-8.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-5+3\)
\(=50-8\)
\(=42\)
\(e)10-\left[\left(8^2-48\right).5+\left(2^3.10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right).5+\left(8.10+8\right)\right]:28\)
\(=10-\left[16.5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6\)
\(=4\)
\(f)8697-\left[3^7:3^5+2.\left(13-3\right)\right]\)
\(=8697-\left[3^2+2.\left(13-3\right)\right]\)
\(=8697-\left[9+2.10\right]\)
\(=8697-9+20\)
\(=8697-29\)
\(=8668\)
\(g)2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5.\left[300-10^2\right]\)
\(=2011+5.\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
\(h)695-\left[200+\left(11-1\right)^2\right]\)
\(=695-\left[200+10^2\right]\)
\(=695-200+100\)
\(=695-300\)
\(=395\)
\(i)129-5\left[29-\left(6-1\right)^2\right]\)
\(=129-5\left[29-5^2\right]\)
\(=129-5\left[29-25\right]\)
\(=129-5.4\)
\(=129-20\)
\(=109\)
= 2015 + 5 x [300-100]
= 2015 + 5 x 200
= 2015 + 1000
= 3015
Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
a, Chia hết cho 3 thì nhóm 2 số thành 1 cặp ; chia hết cho 7 thì nhóm 3 số thành 1 cặp
b, Đề phải là A = 2009.2011
Có :A = 2009.(2010+1) = 2009.2010+2009
= 2009.2010+2010-1 = 2010.(2009+1)-1 = 2010^2-1
Vì 2010^2-1 < 2010^2 = B => A < B
c, A = (3^3)^150 = 27^150
B = (5^2)^150 = 25^150
Vì 27^150 > 25^150 => A > B
k mk nha
bài 1 : thực hiện phép tính
a) 3.52+15.22-26:2
= 3.25 + 15.4 - 26 : 2
= 75 + 60 - 13
= 135 - 13
= 122
b) 20:22+59:58
= 20:4 + 5
= 5 + 5
= 10
c) 100:52+7.32
= 100:25 + 7.9
= 4 + 63
= 67
d) 295-(31-22.5)2
= 295-(31-4.5)2
= 295 - 112
= 295 - 121
= 174
e) (-47)-[(45.24-52.12):14]
= (-47)-[(45.16-25.12):14]
= (-47)-[(720-300):14]
= (-47)-( 420:14 )
= (-47) - 30
= -77
f) (-2011)+5.[300-(17-7)2]
= (-2011)+5.(300-102)
= (-2011)+5.(300-100)
= (-2011)+5.200
= (-2011)+1000
= -1011
g) 5.[29-(6-1)2]-129
= 5.(29-52)-129
= 5.(29-25)-129
= 5.4-129
= 20-129
= -109
Đúng thì tik cái nha ! Thanks nhiều !
2011 + 5.[300 - (17 - 7)2]
= 2011 + 5.(300 - 102)
= 2011 + 5.(300 - 100)
= 2011 + 5.200
= 2011 + 1000
= 3011
bạn gặp đúng người đấy.Kết quả đúng nhất là 3011,nếu sai mình chịu gì cũng được
a) 2017 + 5.[ 300 - \(\left(17-7\right)^2\)]
= 2017 + 5.[ 300 - \(10^2\)]
= 2017 + 5.[ 300 - 100]
= 2017 + 5. 200
= 2017 + 1000
= 3017
b) \(5^{27}\).5.\(5^{25}\)-|-125|
= \(5^{27}\). 5 . \(5^{25}\) - 125
= \(5^{53}\) - 125
= \(5^{53}\) - \(5^3\)
= \(5^{53}\)+ 3
c) (\(5^{25}\).18+ \(5^{15}\).7) : \(5^{17}\)
= [ (\(5^{25}\) . \(5^{15}\)) . ( 18 . 7) ] : \(5^{17}\)
= [ \(5^{40}\) . 126 ] : \(5^{17}\)
= [ \(5^{40}\) : \(5^{17}\) ] . 126
= \(5^{23}\) . 126
Phần c) chưa chắc làm đúng nha
Học tốt :'3
Ta có: \(2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5\left[300-10^2\right]\)
\(=2011+1000=3011\)
\(2011+5.\left[300-\left(17-7\right)^2\right]\)
\(=2011+5.\left[300-\left(10\right)^2\right]\)
\(=2011+5.\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
\(2011+5\left[300-\left(17-7^2\right)\right]\)
\(=2011+5\left[300-\left(17-49\right)\right]\)
\(=2011+5\left[300-\left(-32\right)\right]\)
\(=2011+5.332\)
\(=2011+1660\)
\(=3671\)
\(2011+5\left[300-\left(17-7^2\right)\right]\)
\(=2011+5\left[300-\left(17-49\right)\right]\)
\(=2011+5\left[300-\left(-32\right)\right]\)
\(=2011+5.332\)
\(=2011+1660\)
\(=3671\)