Hiệu điện thế ở 2 đầu dây dẫn:

\(U=I.R=2.10=20\left(V\right)\)

a) \(\Rightarrow2\left(n+3\right)-38⋮\left(n+3\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+3\right)\inƯ\left(38\right)=\left\{19;38\right\}\)

\(\Rightarrow n\in\left\{16;35\right\}\)

b) \(\Rightarrow5\left(n+5\right)-74⋮\left(n+5\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+5\right)\inƯ\left(74\right)=\left\{37;74\right\}\)

\(\Rightarrow N\in\left\{32;69\right\}\)

\(x:0,16=9:x\)

\(\Rightarrow x^2=0,16.9\)

\(\Rightarrow x^2=1,44\)

\(\Rightarrow x=\pm1,2\)

\(a\equiv7\left(mod9\right)\)

\(\Rightarrow a^2\equiv7^2\equiv4\left(mod9\right)\)

Vậy \(a^2\) chia 9 dư 4

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Ta có: a⊥PQ,b⊥PQ

=> a//b

\(\Rightarrow\widehat{N}+\widehat{M}=180^0\)(trong cùng phía)

\(\Rightarrow x=180^0-75^0=105^0\)

Ta có: AB⊥AD,CD⊥AD

=> AB//CD(1)(từ vuông góc đến song song)

Ta có: \(\widehat{CDE}=\widehat{E}=130^0\)

Mà 2 góc này so le trong

=> CD//EF(2)

(1),(2)

=> AB//EF