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Bài 1:
Ta có:
\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)
\(\Leftrightarrow N< M\)
Vậy \(M>N.\)
Bài 2:
Ta có:
\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)
\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)
\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
\(\Leftrightarrow A>B\)
Vậy \(A>B.\)
Bài 3:
\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)
\(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)
\(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)
Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)
\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm
\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)
Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)
Bài 4:
\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)
Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)
\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)
\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)
Vậy \(\frac{1991.1999}{1995.1995}< 1.\)
Bài 4.3.Tính:
a) 27 +1-43
=28-43
=-15
b) |-59| + |-61|
=59+61
=120
c) |126| + |-34|
=126+34
=160
a) 27 +1-43
=27+(-42)
=15
b) |-59| + |-61| ;
= 59+61
=120
c) |126| + |-34|.
=126+34
=160
học tốt
a) 3,54 x 73 + 0,23 x25 +3,54 x 27 +3,54 x 27 +0,17 x 25
= 258,42 + 5,75 + 95,58 + 95,58 + 4,25
= 459,58
b) 563 x 23 +23 x 36 +23
= 12949 + 828 + 23
= 13800
a)Bị lỗi đề bài nhá,nếu là tính nhanh thì vậy,mình xin sửa và làm:
\(3,54\times73+0,23\times25+3,54\times27+0,17\times25\)
\(=3,54\times\left(73+27\right)+25\times\left(0,17+0,23\right)\)
\(=3,54\times100+25\times0,5\)
\(=354+12,5\)
\(=366,5\)
b)
\(563\times23+23\times36+23\)
\(=563\times23+23\times37\)
\(=23\times\left(563+37\right)\)
\(=23\times600\)
\(=13800\)
Cái phần lỗi đề bài đấy mai mình giải không sửa đề.
Không biết nó lỗi thật hay sự cố nhưng mà vẫn : CHÚC BẠN HỌC TỐT
a/3^34=(3^3)^11 x 3
=27^11 x 3
5^20 = (5^2)^10
= 25^10
có 27^11 x3> 25^10(27>25 và 11>10)
suy ra 3^34>5^20
b/17^20=(17^2)^10
=289^10
có 289>71 ; 10>5
nên 71^5>17^20
Toán lớp 6 mà
2727:27+99
=101+99
=200
ai tick cho tớ đầu tiên thì báo tớ tớ tick lại cho!
a) \(\frac{10061006}{20122012}=\frac{10061006}{10061006\times2}=\frac{1}{2}\).
b) \(\frac{a}{35}=\frac{27}{45}\Leftrightarrow a=\frac{27\times35}{45}=\frac{3\times9\times5\times7}{5\times9}=21\).
a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50
=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3
3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)
3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60
3A = -2.3.4 + 49.60.61
\(A=\frac{-2.3.4+49.60.61}{3}=59772\)
b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53
=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6
6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)
6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53
6B = 1.3 + 51.53.55
\(B=\frac{1.3+51.53.55}{6}=24778\)
cau c mk ko bk
d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
D = 30+31+32+33+34+35+36+37+38
=> 3D = 31+32+33+...+38+39
=> 3D - D = 39-30
2D = 39-1
\(D=\frac{3^9-1}{2}=9841\)
27/54>27/55
\(\frac{27}{55}>\frac{27}{54}\)