Question

[Nguyễn Hoàng Quân]

* Bài này hơi khó nên mong thầy phynit cho ai trả lời đúng được 2GP ạ !

Answer 1

[Ngô Tấn Đạt]

2a) \(\frac{1979.1979+1980.21+1958}{1980.1979-1979.1978}=\frac{1979.1979+1979.21+21+1958}{1979\left(1980-1978\right)}=\frac{1979\left(1979+21+1\right)}{1979.2}=\frac{1979.2001}{1979.2}=\frac{2001}{2}\)

b) \(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}=\frac{5^2.6^{11}.\left(2^4\right)^2+6^2.12^6.15^2}{2.6^{12}.2^4.5^4-9^4.960^3}=\frac{5^2.6^{11}.2^8+2^2.3^2.3^6.4^6.3^2.5^2}{2^5.2^{12}.3^{12}.5^4-9^4.960^3}=\frac{5^2.2^{11}.3^{11}.2^8+2^2.3^{10}.2^{12}.5^2}{2^{17}.3^{12}.5^4-3^8.2^6.3.5}=\frac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-3^9.2^6.5}\)

\(\frac{5^2\left(2^{19}.3^{11}+2^{14}.3^{10}\right)}{3^9\left(2^{17}.3^3.5^4-1.2^5.5\right)}=\frac{5^2\left(2^{14}\left(2^5+1\right)+3^{10}\left(3+1\right)\right)}{3^9.\left(2211839840\right)}=\frac{5^2\left(2^{14}.33+3^{10}.4\right)}{3^9.2211839840}=\frac{19421700}{3^9.2211839840}\)

3) a) 6n+99 chia hết cho 3n+4

=> 2(3n+4)+91 chia hết cho 3n+4

=> 91 chia hết cho 3n+4

=> 3n+4 thuộc Ư(91)=1;7;13;91

=> n thuộc 1;3;29

Answer 2

[Nguyễn Hoàng Quân]

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