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18 tháng 9 2021

63=32.7

14=2.7

70=2.5.7

=> ƯCLN (63;14;70) = 7

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a) Các góc kề với \(\widehat {tOz}\)là: \(\widehat {zOy},\widehat {zOn},\widehat {zOm}\)

b) Ta có: \(\widehat {mOn}\) = 30\(^\circ \) nên góc kề bù với \(\widehat {mOn}\) có số đo là: 180\(^\circ \) - 30\(^\circ \) = 150\(^\circ \)

c) Ta có:

\(\begin{array}{l}\widehat {mOn} + \widehat {nOy} + \widehat {yOt} = 180^\circ \\ \Rightarrow 30^\circ  + \widehat {nOy} + 90^\circ  = 180^\circ \\ \Rightarrow \widehat {nOy} = 180^\circ  - 30^\circ  - 90^\circ  = 60^\circ \end{array}\)

Vậy \(\widehat {nOy} = 60^\circ \)

d) Ta có: \(\widehat {tOz} = 45^\circ \) nên góc kề bù với \(\widehat {tOz}\) có số đo là: 180\(^\circ \) - 45\(^\circ \) = 135\(^\circ \)

B(-5)\(\in\){0;5;10}

Ư(-24)={-1;-2;-3;-4;-6;-8;-12;-24;1;2;3;4;6;8;12;24}

Chúc bn học tốt

15 tháng 1 2020

TL:

- B(-5)=(0;-5;-15)

-Ư(-24)=(1;-1;2;-2;3;-3;4;-4;6;-6;8;-8;12;-12;24;-24)

26 tháng 12 2022

đợi tý

18 tháng 8 2023

Đã trả lời rồi còn độ tí đồ ngull

8 tháng 12 2015

a) goi hai so la a ; b va a >b

vi UCLN(a,b)=18=>a=18k            ;       b=18q       (trong do UCLN (k,q)=1 va k>q)

=>a+b=162

18k+18q =162

18(k+q)=162

k+q=9

ta co bang sau   

 

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29 tháng 10 2016

21453 

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8 tháng 2 2018

\(x^2+2x+3\)

\(=\left(x^2+2x+1\right)+2\)

\(=\left(x+1\right)^2+2\)

Do \(\left(x+1\right)^2\ge0\) với mọi x

\(\Rightarrow x^2+2x+3\ge2\)

Dấu = khi x=-1

29 tháng 6 2023

Bài 1:

60= 22.3.5 ; 88 = 23.11

ƯCLN(60;88)= 2= 4

ƯC(60;88)=Ư(4)={1;2;4}

29 tháng 6 2023

Bài 2:

24= 23.3 ; 30=2.3.5 ; 40 = 23.5

BCNN(24;30;40)=23.3.5= 120

BC(24;30;40)=B(120)={0;120;240;360;...}

27 tháng 8 2019

\(C=4,5\cdot\left|2x-0,5\right|-0,25\)

Do \(\left|2x-0,5\right|\ge0\)

=> \(C=4,5\cdot\left|2x-0,5\right|-0,25\ge-0,25\)

Dấu bằng xảy ra khi và chỉ khi \(\left|2x-0,5\right|=0\)hay \(\left|2x-\frac{1}{2}\right|=0\)=> \(2x=\frac{1}{2}\)=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy Cmin = -1/4 khi x = 1/4

\(D=-\left|3x+4,5\right|+0,75\)

Do \(\left|3x+4,5\right|\ge0\)

=> \(-\left|3x+4,5\right|\le0\)

=> \(D=-\left|3x+4,5\right|+0,75\le0,75\)

Dấu bằng xảy ra khi và chỉ khi \(\left|3x+4,5\right|=0\)=> \(\left|3x+\frac{9}{2}\right|=0\)=> \(3x=-\frac{9}{2}\)=> x = \(-\frac{9}{2}:3=\frac{-9}{6}=\frac{-3}{2}\)

Vậy Dmax = 0,75 khi x = -3/2

\(E=\left|x-2005\right|+\left|x-2004\right|\)

\(=\left|x-2005\right|+\left|2004-x\right|\)

\(\ge\left|x-2005+2004-x\right|=\left|-1\right|=1\)

Vậy \(E\ge1\), E đạt giá trị nhỏ nhất là 1 khi \(2004\le x\le2005\)