Hình vẽ:

B

a: ΔABC cân tại A

=>\(\widehat{ABC}=\widehat{ACB}\)

=>\(\widehat{ACB}=47^0\)

ΔABC cân tại A

=>\(\widehat{BAC}=180^0-2\cdot\widehat{ABC}=180^0-2\cdot47^0=86^0\)

b: Xét ΔAMB và ΔAMC có

AM chung

MB=MC

AB=AC

Do đó: ΔAMB=ΔAMC

c Xét ΔAMB có AM+BM>AB

mà AB=AC(ΔABC cân tại A)

nên AM+BM>AC

i: Đặt x²+2x+3=0

=>\(x^2+2x+1+2=0\)

=>\(\left(x+1\right)^2+2=0\)(vô lý)

=>\(x\in\varnothing\)

k: \(3x^2+8x-11=0\)

=>\(3x^2+11x-3x-11=0\)

=>(3x+11)(x-1)=0

=>\(\left[{}\begin{matrix}x=-\dfrac{11}{3}\\x=1\end{matrix}\right.\)

m: Đặt \(x^2+8x+20=0\)

=>\(x^2+8x+16+4=0\)

=>\(\left(x+4\right)^2+4=0\)(vô lý)

=>\(x\in\varnothing\)

Sửa đề: \(\dfrac{1}{2}x+\dfrac{3}{2}\left(x-5\right)=3\)

=>\(\dfrac{1}{2}x+\dfrac{3}{2}x-\dfrac{15}{2}=3\)

=>\(2x=3+\dfrac{15}{2}=\dfrac{21}{2}\)

=>\(x=\dfrac{21}{2}:2=\dfrac{21}{4}\)

\(\dfrac{1}{2}\) x \(\dfrac{3}{2}\) x (\(x-5\)) = 3

         \(x-5\) = 3 : (\(\dfrac{1}{2}\) x \(\dfrac{3}{2}\))

          \(x-5\) = 4

          \(x\)       = 4 + 5

          \(x\)      = 9

Vậy \(x=9\)

a: \(M\left(x\right)=2x^3-x^2+8x-1-3x^2\)

\(=2x^3+\left(-x^2-3x^2\right)+8x-1\)

\(=2x^3-4x^2+8x-1\)

\(N\left(x\right)=-4x^2+2x^2+3+5+7x\)

\(=\left(-4x^2+2x^2\right)+7x+8\)

\(=-2x^2+7x+8\)

b: M(x)+N(x)

\(=2x^3-4x^2+8x-1-2x^2+7x+8\)

\(=2x^3-6x^2+15x+7\)

c: \(3x^2\left(5x^2-x+2\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot x+3x^2\cdot2\)

\(=15x^4-3x^3+6x^2\)