\(A=3,4.10^{-8}\)

\(A=3,4.10.10^{-9}\)

\(A=34.10^{-9}\)

Vậy \(\frac{A}{B}=\frac{34.10^{-9}}{34.10^{-9}}=1\)

Ta có: \(B=34.10^{-9}=3,4.10^{-8}=A\)

                    Vậy A gấp 1 lần B hay A = B

a) Ta có: B=34.10\(^{-9}\)=34.\(\frac{1}{10^9}\)=34.\(\frac{1}{10}\).\(\frac{1}{10^8}\)=3,4.10\(^{-8}\)

Vậy A=B

b) Ta có: A=10\(^{-4}\)+10\(^{-3}\)+10\(^{-2}\)=0,0111

B=10\(^{-9}\)=\(\frac{1}{10^9}\)=\(\frac{1}{1000000000}\)=0,000000001

\(\Rightarrow\)A= 11100000B

Câu a)
\(\hept{\begin{cases}A=3,4\cdot10^{-8}=3,4\cdot\frac{1}{10^8}=\frac{3,4}{10^8}\\B=34\cdot10^{-9}=34\cdot\frac{1}{10^9}=\frac{3,4\cdot10}{10^8\cdot10}=\frac{3,4}{10^8}\end{cases}}\)
\(\Rightarrow\frac{A}{B}=1\)

Câu b)
\(\hept{\begin{cases}A=10^{-4}+10^{-3}+10^{-2}=\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}=\frac{10^5+10^6+10^7}{10^9}\\B=10^{-9}=\frac{1}{10^9}\end{cases}}\)
\(\Rightarrow\frac{A}{B}=10^5+10^6+10^7=11100000\)

a ) A gấp B 10 lần

Trả lời  : 

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Câu hỏi của Uzumaki Naruto - Toán lớp 7 - Học toán với OnlineMath 

Chúc bn hc tốt <3 

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https://olm.vn/hoi-dap/detail/54407333464.html

Tớ bt lm vậy cậu sẽ ko thấy 

a) A = 3,4 . 10^-8 = \(\frac{3,4}{10^{-8}}=\frac{34}{10^{-9}}\)

B = 34 . 10^-9 = \(\frac{34}{10^{-9}}\)

A = B

b) \(\frac{A}{B}=\frac{\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}}{\frac{1}{10^9}}\)

\(=\left(\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}\right).10^9\)

\(=\frac{1+10+10^2}{10^4}.10^9\)

\(=111.10^5\)

=> A = 11100000B

a)1 lần

b)11100000 lần

a) 1 lần

A=3.4.\(\frac{1}{10^8}\)=\(\frac{12}{10^8}\)

B=34.\(\frac{1}{10^9}\)=\(\frac{34}{10^9}\)

Ta có: A=k.B\(\Rightarrow\)\(\frac{12}{10^8}=k.\frac{34}{10^9}\)=>k=\(\frac{12}{10^8}:\frac{34}{10^9}\)=\(\frac{12}{10^8}.\frac{10^9}{34}\)=\(\frac{60}{17}\)