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Bài 1:
a. https://olm.vn/hoi-dap/detail/100987610050.html
b. Giống nhau hoàn toàn => P=Q
Chỉ biết thế thôi
a) \(4.5^2-32:2^5\)
\(=4.25-2^5:2^5\)
\(=100-1\)
\(=99.\)
b) \(9.8.14+6.\left(-17\right)\left(-12\right)+19.\left(-4\right).18\)
\(=9.2.4.14+6.3.\left(-4\right)\left(-17\right)+76.18\)
\(=18.56+18.68+18.76\)
\(=18\left(56+68+76\right)\)
\(=18\left(132+68\right)\)
\(=18.200\)
\(=3600.\)
c) \(\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\left(\dfrac{-1}{2}\right)^2+2.\dfrac{-1}{2}+3\right]+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+\left(-1\right)+3\right]+1\)
\(\)\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+2\right]+1\)
\(=\left(\dfrac{-1}{2}\right).\dfrac{9}{4}+1\)
\(=\dfrac{-9}{8}+1\)
\(=\dfrac{-1}{8}\)
\(A=\dfrac{12n+1}{30n+2}\)
Gọi \(d\)là \(UCLN\left(12n+1;30n+2\right)\)
\(\left\{{}\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}60n+5⋮d\\60n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)
\(1⋮d\Rightarrow d=1\)
Vậy phân số trên tối giản
b tương tự
1,Gọi a là ƯCLN(12n+1;30n+2).Nên ta có:
12n+1 chia hết cho d và 30n+2 chia hết cho d
<=>5.(12n+1) chia hết cho d và 2.(30n+2) chia hết cho d
<=>60n+5 chia hết cho d và 60n+4 chia hết cho d
=>(60n+5)-(60n+4) chia hết cho d
=>1 chia hết cho d =>d = 1
Vậy d=1 =>\(\dfrac{12n+1}{30n+2}\) là phân số tối giảm (đpcm )
b1: a, 612.(15+19-34)=612.0=0
b,414.(37.4+23.4-240)=414.0=0
c,(517.125-518.25)+63:23=(517.53-518.52)+33=0+27=27
b2:a,143+7.(n-17)=206
===> 7.(n-17)=206-143=63
====>n-17=63:7=9
=====>n=9+17=26
vậy n=26
b,128-28:(15-n)=124
====>28:(15-n)=128-124=4
=====> 15-n=28:4=7
=====> n=15-7=8
vậy n=8
c,3n.2+48=210
====>3n.2=210-48=162
====>3n=162:2=81=34
====>n=4
vậy n=4
Gọi ƯCLN (12n+1,30n+2) là d
\(\Rightarrow\left(12n+1\right)⋮d\)
\(\left(30n+2\right)⋮d\)
\(\Rightarrow5\left(12n+1\right)-2\left(30n+2\right)⋮d\)
\(\Rightarrow60n+5-60n-4⋮d\)
\(\Rightarrow1⋮d\Leftrightarrow d=1\)
Vậy ƯCLN \(\left(12n+1,30n+2\right)=1\Leftrightarrow\dfrac{12n+1}{30n+2}\) là p/s tối giản \(\left(dpcm\right)\)
Gọi ước chung lớn nhất của 12n+1 và 30n+ 2 là d
\(\Rightarrow\) ( 12n+1) \(⋮\) d và ( 30n+2 ) \(⋮\) d
\(\Rightarrow\) \(\left[5\left(12n+1\right)-2\left(30n+2\right)\right]⋮d\)
\(\Leftrightarrow\) ( 60n + 5 - 60n - 4 ) \(⋮d\)
\(\Leftrightarrow\) 1 \(⋮\) d hay d= 1
Vậy ước chung lớn nhất của 12n+ 1 và 30n+2 là 1 hay \(\dfrac{12n+1}{30n+2}\) là phân số tối giản .
Câu 2:
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\)
\(=2014\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\)
\(=2014\left(1+\dfrac{1}{2\left(2+1\right)}.2+\dfrac{1}{3\left(3+1\right)}.2+...+\dfrac{1}{2013\left(2013+1\right)}.2\right)\)
\(=2014\left(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2013.2014}\right)\)
\(=4028\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)
Bạn tự tính nốt nhé
1)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\left(1\right)\)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\\ =\dfrac{1}{1}-\dfrac{1}{2012}< 1\left(2\right)\)
Từ (1) và (2) ta có: A < 1
2)
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\\ =2014\cdot\left(\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\\ =2014\cdot\left(\dfrac{1}{\left(1\cdot2\right):2}+\dfrac{1}{\left(2\cdot3\right):2}+\dfrac{1}{\left(3\cdot4\right):2}+...+\dfrac{1}{\left(2013\cdot2014\right):2}\right)\\ =2014\cdot\left(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{2013\cdot2014}\right)\\ =2014\cdot2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2013\cdot2014}\right)\\ =4028\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\\ =4028\cdot\left(1-\dfrac{1}{2014}\right)\\ =4028\cdot\dfrac{2013}{2014}\\ =4026\)
3)
Để A là số nguyên thì \(6n+42⋮6n\Rightarrow42⋮6n\Rightarrow6n\inƯ\left(42\right)\)
\(Ư\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)
| 6n | 1 | 2 | 3 | 6 | 7 | 14 | 21 | 42 |
| n | \(\dfrac{1}{6}\) | \(\dfrac{1}{3}\) | \(\dfrac{1}{2}\) | 1 | \(\dfrac{7}{6}\) | \(\dfrac{7}{3}\) | \(\dfrac{7}{2}\) | 7 |
Vì n là số tự nhiên nên n = 1 hoặc n = 7
4)
\(A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)
Vậy A<B
Bài 1.
Đặt (12n + 1; 30n + 2) = d
\(\Rightarrow\) \(\left\{{}\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}60n+5⋮d\\60+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) (60n + 5) - (60n + 4) \(⋮\) d
\(\Rightarrow\) 1 \(⋮\) d
\(\Rightarrow\) d = 1
\(\Rightarrow\) (12n + 1; 30n + 2) = 1
Vậy phân số \(\dfrac{12n+1}{30n+2}\) là phân số tối giản