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\(M=\left\{1;3;5;6;10;15;30\right\}\)
\(P=\left\{21;24;27;30;33;36;39;42;45;48\right\}\)
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Câu 2:
(x+5)(3x-12)>0
=>(x-4)(x+5)>0
=>x>4 hoặc x<-5
Vậy: S=Z\[-5;4]
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a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)
\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)
\(=31x+465=620\)
\(=31x=620-465\)
\(=31x=155\)
\(=x=155\div31\)
\(x=5\)
b) \(2+4+6+8+....+2x = 210\)
\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)
\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)
\(\Rightarrow2+4+6+8+x=210\div2\)
\(\Rightarrow2+4+6+8+...+x=105\)
\(\Rightarrow x=14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
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Bài 5:Giải:
Ta có: \(\left\{{}\begin{matrix}a+3c=2016\left(1\right)\\a+2b=2017\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow a=2016-3c\)
Lấy \(\left(2\right)-\left(1\right)\) ta được:
\(2b-3c=1\Leftrightarrow b=\dfrac{1+3c}{2}\)
Khi đó:
\(P=a+b+c=\left(2016-3c\right)+\dfrac{1+3c}{2}\) \(+\) \(c\)
\(=\left(2016+\dfrac{1}{2}\right)+\dfrac{-6c+3c+2c}{2}\)
\(=2016\dfrac{1}{2}-\dfrac{c}{2}\) Vì \(a,b,c\ge0\) nên:
\(P=2016\dfrac{1}{2}-\dfrac{c}{2}\le2016\dfrac{1}{2}\)
Vậy \(P_{max}=2016\dfrac{1}{2}\Leftrightarrow c=0\)
1)x=1;2;3;5;6;10;15;30
2)x=rỗng
3)x=0;7