Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

AB<AC

=>góc B>góc C

=>90 độ-góc B<90 độ-góc C

=>góc HAB<góc HAC

\(AB< AC\\ \Rightarrow\widehat{C}< \widehat{B}\)

Xét tam giác \(AHB\) và \(AHC\) có

\(\left\{{}\begin{matrix}\widehat{HAB}=90^o-\widehat{B}\\\widehat{HAC}=90^o-\widehat{C}\end{matrix}\right.\)

mà \(\widehat{B}>\widehat{C}\)

\(\Rightarrow\widehat{HAB}< \widehat{HAC}\)

a: =>x=75:25%=300

b: =>11/12x=-7/12

hay x=-7/11

c: \(\Leftrightarrow\left(x-1\right)\cdot\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{3}{4}=\dfrac{-1}{4}\)

=>x-1=-1

hay x=0

\(a,=0,25.x=75\Rightarrow x=75:0,25=300\)

\(b,\dfrac{11}{12}x=\dfrac{1}{6}-\dfrac{3}{4}=\dfrac{-7}{12}\)

\(x=\dfrac{-7}{12}:\dfrac{11}{12}=\dfrac{-7}{11}\)

\(c,\dfrac{1}{4}\left(x-1\right)=\dfrac{1}{2}-\dfrac{3}{4}=\dfrac{-1}{4}\)

\(x-1=\dfrac{-1}{4}:\dfrac{1}{4}=-1\)

\(\Leftrightarrow x=0\)

a: Xét ΔHAC vuông tại H và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔHAC~ΔABC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=15^2+20^2=625\)

=>BC=25

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=BA^2\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH\cdot25=15^2=225\\AH\cdot25=15\cdot20=300\end{matrix}\right.\)

=>BH=9; AH=12

Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)

Ta có: xy=112

\(\Leftrightarrow28k^2=112\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\cdot2=8\\y=7k=7\cdot2=14\end{matrix}\right.\)

Trường hợp 2: x=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-8\\y=7k=-14\end{matrix}\right.\)

14.D

15.B

16.C

d b c