giúp mk, mk tick cho
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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(-\(\frac{4}{12}\)) +7=(\(-\frac{1}{3}\)) + 7
=(-\(\frac{1}{3}\))+\(\frac{21}{3}\)
=\(\frac{20}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`c)A=(9^4*27^5*3^6*3^5)/(3^8*81^4*234*8^2)`
`A=(3^8*3^15*3^11)/(3^8*3^16*3^2*2*13*2^6)`
`A=(3^34)/(3^36*2^8*13)`
`A=1/(3^2*2^8*13)`
`B=(4^6*9^5+6^9.120)/(8^4*3^12-6^11)`
`B=(2^12*3^15+2^12*3^10*5)/(2^12*3^12-2^11*3^11)`
`B=(2^12*3^10(3^5+5))/(2^11*3^11*(6-1))`
`B=(2^12*3^10*248)/(2^11*3^11*5)`
`B=(248*2)/(3*5)`
`B=496/15`
c) A = \(\dfrac{9^4.27^5.3^6.3^5}{3^8.81^4.234.8^2}=\dfrac{3^8.3^{15}.3^6.3^5}{3^8.3^{16}.2.3^2.13.2^6}=\dfrac{3^{34}}{3^{26}.2^7.13}=\dfrac{3^8}{2^7.13}\)
d) B =\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta AHB=\Delta AHC\Rightarrow\widehat{AHB}=\widehat{AHC}\) (2 góc tương ứng)
Mà \(\widehat{AHB}+\widehat{AHC}=180^o\) (2 góc kề bù)
\(\Rightarrow\widehat{AHB}=\widehat{AHC}=90^o\)
\(\Rightarrow AH\perp BC\left(đpcm\right)\)
\(\Delta AHB=\Delta AHC\Rightarrow\widehat{AHB}=\widehat{AHC}\left(2.góc.tương.ứng\right)\)
Ta có:\(\widehat{AHB}+\widehat{AHC}=180^o\left(2.góc.kề.bù\right)\\ \Rightarrow2\widehat{AHB}=180^o\left(vì\widehat{AHB}=\widehat{AHC}\right)\\ \Rightarrow\widehat{AHB}=90^o\\ \Rightarrow AH\perp BC\)
b) ĐKXĐ: \(n\ne-\dfrac{1}{3}\left(đúng.do.n\in Z\right)\)
\(\dfrac{6n-2}{3n+1}=2-\dfrac{4}{3n+1}\in Z\)
\(\Rightarrow\left(3n+1\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
Do \(n\in Z\Rightarrow n\in\left\{-1;0;1\right\}\)
c) ĐKXĐ: \(n\ne2\)
\(\dfrac{n^2-3n+5}{n-2}=\dfrac{\left(n-2\right)\left(n-1\right)+3}{n-2}=n-1+\dfrac{3}{n-2}\in Z\)
\(\Rightarrow\left(n-2\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)