x x 2 - 25 - x - 5 x 2 + 5 x : 2 x - 5 x 2 + 5 x + x 5 - x
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`1, -2/9 xx 15/17 + (-2/9) xx 2/17`
`= -2/9 xx (15/17 + 2/17)`
`= -2/9 xx 17/17`
`=-2/9xx1`
`=-2/9`
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`-5/3 xx 6/5 + (-7/9) xx 3/10`
`= -30/15 + (-21/90)`
`= -2 + (-7/30)`
`=-60/30 +(-7/30)`
`=-67/30`
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`15/20 xx 7/5 + (-9/7) xx (-6/4)`
`=3/4 xx7/5 + (-9/7) xx(-6/4)`
`= 21/20 + 54/28`
`= 21/20 + 27/14`
`=417/140`
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`-25/13 xx 5/19 + (-25/13) xx 14/19`
`=-25/13 xx (5/19 +14/19)`
`=-25/13 xx 19/19`
`= -25/13 xx 1`
`=-25/13`
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`-7/13 xx 13/5 + (-9/7) xx 5/3`
`=-7/5 +(-15/7)`
`=-124/35`
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
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a: =>(x-2)^3*[(x-2)^8-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{2;3;1\right\}\)
b: (x-5)^24=(x-5)^9
=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)
=>x-5=0 hoặc x-5=1
=>x=6 hoặc x=5
c: =>(x-5)^4*[(x-5)^21-1]=0
=>x-5=0 hoặc x-5=1
=>x=5 hoặc x=6
a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)
\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
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1. 444 x 5 = 222 x 2 x 5 = 222 x 10 = 2220
a.Đúng b.Sai
2. 282 x 5 = 280 + 2 x 5 = 280 x 10 = 2800
a.Đúng b. Sai
3. 4 x 8 x 7 x 25 = (8 x 7) x (25 x4) = 56 x 100 = 5600
a.Đúng b.Sai
4. 25 x 8 x 9 = (25 x 4) x (4 x 9) = 100 x 36 = 3600
a.Đúng b.Sai