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3:
a:Các tia trên hình là Ax,Ay,Bx,By,Cx,Cy
=>Có 6 tia
b: AB<AC
=>B nằm giữa A và C
=>AB+BC=AC
=>BC=4cm
c: AI=3/2=1,5cm
CI=7-1,5=5,5cm
![](https://rs.olm.vn/images/avt/0.png?1311)
x + 6 chia hết cho x + 3
=> x + 3 + 3 chia hết cho x + 3
=> 3 chia hết cho x + 3
=> (x + 3) \(\in\) Ư(3)
=> (x + 3) \(\in\) {-3; -1; 1; 3}
=> x \(\in\) {-6; -4; -2; 0}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
\(\Leftrightarrow6n+4⋮2n-1\)
\(\Leftrightarrow2n-1\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{1;0;4;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`1a)-17/30-11/(-15)+(-14)/24`
`=-17/30+22/30+(-7)/12`
`=5/30+(-7)/12`
`=1/6-7/12=2/12-7/12=-5/12`
`1b)(-10)/11*4/7+(-10)/11*3/7+1 10/11`
`=(-10)/11*(4/7+3/7)+1+10/11`
`=-10/11+10/11+1=1`
`1c)(5/7*0,6-5:3 1/2).(40%-1,4).(-2)^3`
`=(5/7*3/5-5:7/2).(0,4-1,4).(-8)`
`=(3/7-10/7).(-1).(-8)`
`=8.(-1)=-8`
![](https://rs.olm.vn/images/avt/0.png?1311)
Ũa z hỏi lm j k bik lm mới hỏi chứ sắp hết thời gian r mà k lm đc=))
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\(=\dfrac{2^{19}\cdot3^9\cdot5+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot5\left(2+1\right)}{2^{19}\cdot3^9\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{5\cdot3}{-5}=-\dfrac{3}{2}\)
\(A=\dfrac{2^{19}\cdot27^3\cdot5-15\cdot\left(-4\right)^9\cdot9^4}{6^9\cdot2^{10}-\left(-12\right)^{10}}\)
\(A=\dfrac{2^{19}\cdot\left(3^3\right)^3\cdot5-3\cdot5\cdot-\left(2^2\right)^9\cdot\left(3^2\right)^4}{2^9\cdot3^9\cdot2^{10}-\left(2^2\right)^{10}\cdot3^{10}}\)
\(A=\dfrac{2^{19}\cdot3^9\cdot5+3^9\cdot2^{18}\cdot5}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(A=\dfrac{2^{18}\cdot3^9\cdot5\cdot\left(2+1\right)}{2^{19}\cdot3^9\cdot\left(1-2\cdot3\right)}\)
\(A=\dfrac{1\cdot1\cdot5\cdot3}{2\cdot1\cdot-5}\)
\(A=-\dfrac{1}{2}\cdot3\)
\(A=-\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(=\dfrac{-27\cdot100}{-30}=\dfrac{2700}{30}=90\)
\(a.2^{x+2}-2^x=96\\ 2^x\cdot4-2^x=96\\ 2^x\cdot\left(4-1\right)=96\\ 2^x\cdot3=96\\ 2^x=96:3\\ 2^x=32\\ 2^x=2^5\\ =>x=5\)
\(2^{x+1}\cdot3^y=12^x\\ 2^{x+1}\cdot3^y=3^x\cdot2^{2x}\\ \left\{{}\begin{matrix}2^{x+1}=2^{2x}\\3^y=3^x\end{matrix}\right.\\ \left\{{}\begin{matrix}x+1=2x\\y=x\end{matrix}\right.\\ \left\{{}\begin{matrix}x=1\\y=x=1\end{matrix}\right.\\ \left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(c.10^x:5^y=20^y\\ 10^x:5^y=5^y\cdot4^y\\ 10^x=10^{2y}\\ =>x=2y\)