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Ta có : \(m=m_1+m_2< =>m_1+m_2=644\)

\(< =>m_2=644-m_1\left(+\right)\)

\(V=V_1+V_2< =>\frac{m}{D}=\frac{m_1}{D_1}+\frac{m_2}{D_2}\)

\(< =>\frac{644}{8,3}=\frac{m_1}{7,3}+\frac{m_2}{11,3}\left(++\right)\)

Thế \(\left(+\right)\)vào \(\left(++\right)\)ta được : 

\(\frac{644}{8,3}=\frac{m_1}{7,3}+\frac{644-m_1}{11,3}\)

giải phương trình trên ta được : \(m_1=438\left(g\right)\)

Mặt khác : \(m_2=m-m_1=644-438=226\left(g\right)\)

Vậy ... 

Đặt \(A=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}\)

\(A=\frac{3^2\cdot\left(2^2\right)^2\cdot2^{32}}{11\cdot2^{13}\cdot\left(2^2\right)^{11}-16^9}\)

\(A=\frac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{13}\cdot2^{22}-16^9}=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-\left(2^4\right)^9}\)

\(A=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}\)

\(A=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2\cdot2^{35}}\)

\(A=\frac{3^2\cdot2^{36}}{\left(11-2\right)\cdot2^{35}}=\frac{9\cdot2^{36}}{9\cdot2^{35}}=\frac{2^{36}}{2^{35}}=2\)

Bài làm:

\(\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

\(=\frac{3^2.2^{36}}{\left(11-2\right).2^{35}}\)

\(=\frac{9.2^{36}}{9.2^{35}}\)

\(=2\)

Học tốt 

Bạn tham khảo: Câu hỏi của Nguyễn Ngọc Ánh - Toán lớp 6 - Học toán với OnlineMath

1/ can

2/ couldn't

3/ won't be able to 

4/ can't

5/ couldn't

6/ could

7/ can

8/ can

9/ can't

Ta thấy : \(\frac{5}{11}>\frac{5}{12}>\frac{5}{13}>\frac{5}{14}\)

 \(S=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}< \frac{5}{11}\times4=\frac{20}{11}< 2\)  (1)

\(S=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}>\frac{5}{14}\times4=\frac{10}{7}>1\)   (2)

Từ (1) và (2) suy ra : \(1< S< 2\)  (ĐPCM)

Sửa đề : Chứng minh : S > 1

Ta thấy : \(\frac{5}{20}>\frac{5}{21}>\frac{5}{22}>\frac{5}{23}>\frac{5}{24}\)

\(\Rightarrow S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>\frac{5}{24}\times5=\frac{25}{24}>1\)

Vậy S > 1 (ĐPCM)