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a) \(199\left(15-17\right)-199\left(-17+5\right)\)
\(=199.15-199.17+199.17-199.5\)
\(=199.\left(15-5\right)\)
\(=199.10=1990\)
b) \(73\left(8-59\right)-59\left(8-73\right)\)
\(=73.8-73.59-59.8+59.73\)
\(=8\left(73-59\right)\)
\(=8.14=112\)
a)2.18.14+3.17.12+4.69.9
=(2.18).14+(12.3).17+(4.9).69
=36.14+36.17+36.69
=36.(14+17+69)
=36.100
=3600
b)1+(-3)+5+(-7)+9+(-11)+13+(-15)
=[1+(-3)]+[5+(-7)]+[9+(-11)]+[13+(-15)]
=(-2)+(-2)+(-2)+(-2)
=(-8)
a)5.(-8).2.(-3)
=[5.(-2)].[(-8).(-3)]
=-10.24
=-240
b)(45-135+72)-(45+72)
=45-135+72-45-72
=(45-45)+(72-72)-135
=0+0-135
=-135
c)3.(-5)2.2+2.(-5)-20
=3.25.2+2(-5)-20
=75.2+(-10)-20
=150+(-30)
=120
d)86.(-46)+46.27-46.41
=86.(-1).46+46.27-46.41
=-86.46+46.27-46.41
=46(-86+27-41)
=46(-100)
-4600
e)34(15-10)-15(34-10)
=34.15-34.10-15.34+15.10
=(34.15-15.34)+(15.10-34.10)
=[34(15-15)]+[10(15-34)]
=34.0+10(-9)
=0+(-90)
=-90
Bài 2:
Ta thấy: 52 > 4.5
62 > 5.6
72 > 6.7
....
20172 > 2016.2017
\(\Rightarrow\)\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
....
\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)
Cộng vế với nhau, ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}\)( vì \(\frac{1}{2017}>0\))
k giúp mik ✅
Mk thấy bài 1 và 2 dễ nên bạn tự làm nha
3
+)Ta có n-2 \(⋮\)n-2
=>2.(n-2)\(⋮\)n-2
=>2n-4\(⋮\)n-2(1)
+)Theo bài ta có:2n+1\(⋮\)n-2(2)
+)Từ (1) và (2)
=>(2n+1)-(2n-4)\(⋮\)n-2
=>2n+1-2n+4\(⋮\)n-2
=>5\(⋮\)n-2
=>n-2\(\in\)Ư(5)={\(\pm\)1;\(\pm\)5}
+)Ta có bảng:
| n-2 | -1 | 1 | -5 | 5 |
| n | 1\(\in\)Z | 3\(\in\)Z | -3\(\in\)Z | 7\(\in\)Z |
Vậy n\(\in\){1;3;-3;7}
Chúc bn học tốt
a. 5.(–8).( –2).(–3) b. 4.(–5)2 + 2.(–5) – 20
=(-5).8.(-2).(-3) ={(-5).2} {4+1}-20
=(-5)(-2)(-3).8 =(-10).5-20=-50-20=-70
=10.(-24)=-240
Lời giải:
a.
$\frac{6}{13}+\frac{-15}{39}=\frac{6}{13}+\frac{-5}{13}=\frac{6+(-5)}{13}=\frac{1}{13}$
b.
$\frac{-8}{18}+\frac{-15}{27}=\frac{-4}{9}+\frac{-5}{9}=\frac{-4-5}{9}=\frac{-9}{9}=-1$
a: \(-\dfrac{15}{8}\left(\dfrac{8}{-15}+\dfrac{32}{27}\right)-\dfrac{15}{7}\)
\(=\dfrac{15}{8}\cdot\dfrac{8}{15}-\dfrac{15}{8}\cdot\dfrac{32}{27}-\dfrac{15}{7}\)
\(=1-\dfrac{15}{7}-\dfrac{15}{27}\cdot\dfrac{32}{8}\)
\(=\dfrac{-8}{7}-4\cdot\dfrac{5}{9}=\dfrac{-8}{7}-\dfrac{20}{9}\)
\(=-\dfrac{72}{63}-\dfrac{140}{63}=-\dfrac{212}{63}\)
b: \(41,54+23,17+8,46-3,17\)
=(41,54+8,46)+(23,17-3,17)
=50+20
=70
c: \(3\cdot\left(32,1-6,32\right)+7\cdot32,1+3\cdot0,32\)
\(=3\cdot32,1-3\cdot6,32+7\cdot32,1+3\cdot0,32\)
\(=32,1\left(3+7\right)-3\left(6,32-0,32\right)\)
\(=321-3\cdot6=321-18=303\)
a; - \(\dfrac{15}{8}\).(\(\dfrac{8}{-15}\) + \(\dfrac{32}{27}\)) - \(\dfrac{15}{7}\)
= - \(\dfrac{15}{18}\).(\(\dfrac{8}{-15}\)) - \(\dfrac{15}{8}\).(\(\dfrac{32}{27}\)) - \(\dfrac{15}{7}\)
= 1 - \(\dfrac{20}{9}\) - \(\dfrac{15}{7}\)
= \(\dfrac{-11}{9}\) - \(\dfrac{15}{7}\)
= - \(\dfrac{212}{63}\)