![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(g,4=2^2;6=2.3\\ \Rightarrow BCNN\left(4,6\right)=2^2.3=12\\ \Rightarrow x\in BC\left(4,6\right)=B\left(12\right)=\left\{0;12;24;36;48;60;...\right\}\\ \text{Mà }0< x< 50\\ \Rightarrow x\in\left\{12;24;36;48\right\}\\ h,12=2^2.3;18=2.3^2\\ \Rightarrow BCNN\left(12,18\right)=2^2.3^2=36\\ \Rightarrow x\in BC\left(12,18\right)=B\left(36\right)=\left\{0;36;72;108;144;180;216;252;...\right\}\\ \text{Mà }x< 250\\ \Rightarrow x\in\left\{0;36;72;108;144;180;216\right\}\)
g,\(x⋮4,x⋮6\Rightarrow x\in BC\left(4,6\right)=\left\{\pm0;\pm12;\pm24;\pm36;\pm48;\pm60;...\right\}\)
Mà \(0< x< 50\Rightarrow x\in\left\{12;36;48\right\}\)
h,\(x⋮12,x⋮18\Rightarrow x\in BC\left(12,18\right)=\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216;\pm252;...\right\}\)
Mà \(x< 50\Rightarrow x\in\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x-1)(y+4)=11
Ta có:11=1.11=11.1=(-1).(-11)=(-11).(-1)
Do đó ta có bảng sau:
y+4 | -11 | -1 | 1 | 11 |
2x-1 | -1 | -11 | 11 | 1 |
2x | 0 | -10 | 12 | 2 |
x | 0 | -5 | 6 | 1 |
y | -15 | -5 | -3 | 7 |
\(\left(2x-1\right)\left(y-2\right)=-11\)
\(2x-1;y-2\inƯ\left(-11\right)=\left\{\pm1;\pm11\right\}\)
Tự lập bảng
![](https://rs.olm.vn/images/avt/0.png?1311)
x + 3 + 9 chia hết x + 3
9 chia hết x + 3
x + 3 thuộc Ư ( 9 )
mà Ư (9) = ( 1,3,9 )
hay x + 3 thuộc ( 1,3,9 )
ta có bảng
x + 3 1 3 9
x -2 0 6
ĐG Loại TM TM
Vậy x thuộc ( 0 , 6 )
![](https://rs.olm.vn/images/avt/0.png?1311)
e: Ta có: \(\left(x+1\right)\left(x+2\right)=444222\)
\(\Leftrightarrow x^2+3x-444220=0\)
\(\text{Δ}=3^2-4\cdot1\cdot\left(-444220\right)=1776889\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{-3-1333}{2}=-668\\x_2=\dfrac{-3+1333}{2}=665\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`
`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`
`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`
`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`
`S_1 = 5/3 (1 - 1/100)`
`S_1 = 5/3 . 99/100`
`S_1 = 33/20`