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\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
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A = \(\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
=\(\dfrac{5}{9}\cdot\dfrac{5}{7}=\dfrac{25}{63}\)
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a) (x-3)(y+2) = 7
=> (x−3)∈Ư(7);(y+2)∈Ư(7)(x−3)∈Ư(7);(y+2)∈Ư(7)
=> (x−3)∈{−7;−1;1;7}(x−3)∈{−7;−1;1;7}
(y+2)∈{−7;−1;1;7}
ta có bảng sau :
vạy có 4 cạp
xin like
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Bài 1 :
\(A=3^0+3^1+3^2+3^3+...+3^{98}\)
\(A=\left(1+3+3^2\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\) ( Nhóm 3 số 1 nhé )
\(A=13+.....+3^{97}.13⋮13\left(\text{đ}pcm\right)\)
Bài 2 :
Theo ý a ta có :
\(A=13+.....+3^{97}.13+3^{99}+3^{100}\)
\(A=13+.....+3^{97}.13+3^{99}.4⋮̸13\)
Bài 3 :
Để D chia hết cho 2 thì x chia hết cho 2
1. \(A=3^0+3^1+3^2+...+3^{98}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{96}\right)\)chia hết cho \(13\).
2. \(B=3^0+3^1+3^2+3^3+...+3^{100}\)
\(=1+3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=4+13\left(3^2+3^5+...+3^{98}\right)\)không chia hết cho \(13\).
3. \(D=\left(12.3+26.b+2022.c+x\right)\)chia hết cho \(2\)
\(\Leftrightarrow x⋮2\)(vì \(12.3⋮2,26b⋮2,2022c⋮2\))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Rightarrow-7< x\le-3\)
Nếu x ∈ Z thì:
\(x\in\left\{-6;-5;-4;-3\right\}\)
\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Leftrightarrow-4< x\le-3\)
Nếu x nguyên thì x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)
\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)
\(\Leftrightarrow\)\(50-6.x=20\)
\(\Leftrightarrow\)\(6.x=50-20\)
\(\Leftrightarrow\)\(6.x=30\)
\(\Leftrightarrow\)\(x=5\)
\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)
\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)
\(\Leftrightarrow\)\(100x+5050=7450\)
\(\Leftrightarrow\)\(100x=7450-5050\)
\(\Leftrightarrow\)\(100x=2400\)
\(\Leftrightarrow\)\(x=24\)
b.
(x+1)+(x+2)+...+(x+100)=7450
=> 100x + (1+2+3+...+100)=7450
=>100x + (100+1).50=7450
=>100x=2400
=>x=24
`1 / 3 + 2 / 3 : x = -7`
`2 / 3 : x = -7 - 1 / 3`
`2 / 3 : x = -22 / 3`
`x = 2 / 3 : -22 / 3`
`x = -1 / 11`
\(\dfrac{1}{2}+\dfrac{2}{3}:x=-7\)
\(\dfrac{2}{3}:x=\left(-7\right)-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{22}{3}\right)\)
\(x=\dfrac{2}{3}\cdot\left(-\dfrac{3}{22}\right)\)
\(x=-\dfrac{6}{66}\)
\(x=-\dfrac{1}{11}\)