1/ \(S_{ΔABC}=\dfrac{1}{2}.AB.AC=\dfrac{3}{2}.AC=8\\↔AC=\dfrac{16}{3}\\ΔABC\backsim ΔMNP\\→\dfrac{AB}{AC}=\dfrac{MN}{MP}=k=\dfrac{3}{\dfrac{16}{3}}=\dfrac{9}{16}\\→D\)

2/ \(|2x-1|=3\\↔2x-1=3\quad or\quad 2x-1=-3\\↔x=2\quad or\quad x=-1\)

$→$ Tổng các nghiệm của pt là 2-1=1

$→ A$

3/ Xét \(ΔBNM\) và \(ΔBCA\):

\(\widehat B:chung\)

\(\dfrac{MA}{MB}=\dfrac{NC}{NB}(gt)\)

\(→ΔBNM\backsim ΔBCA(c-g-c)\)

\(→\dfrac{MN}{AC}=\dfrac{BN}{BC}\)

\(\dfrac{NC}{NB}=\dfrac{2}{5}→NC=\dfrac{2NB}{5}\\→BC=\dfrac{2NB}{5}+NB=\dfrac{7NB}{5}\\↔\dfrac{NB}{BC}=\dfrac{5}{7}\\→\dfrac{MN}{AC}=\dfrac{15}{AC}=\dfrac{5}{7}\\→AC=21(cm)\\→A\)

4/ \(\dfrac{x^2+4x+4}{x+2}(x\ne 2)\\=\dfrac{(x+2)^2}{x+2}=x+2\\→B\)

5/ \( x^3-6x^2y+12xy^2-8y^3\\=x^3-3x^2.2y+3.x.(2y)^2-(2y)^3\\=(x-2y)^3\\→D\)

6/ \(ΔMNP\backsim ΔABC\\→\dfrac{MN}{AB}=\dfrac{NP}{BC}=\dfrac{1}{2}\\→MN=\dfrac{AB}{2}\)

Kẻ đường cao \(AH\) ứng \(BC\)

Áp dụng định lý Pytago vào \(ΔABH\) vuông tại \(H\)

\(→AH=\sqrt{AB^2-BH^2}=\sqrt{AB^2-\dfrac{BC^2}{4}}=\sqrt{AB^2-\dfrac{AB^2}{4}}=\sqrt{\dfrac{3AB^2}{4}}=\dfrac{\sqrt 3AB}{2}(cm)\\S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.\dfrac{\sqrt 3 AB}{2}.AB=4\sqrt 3\\↔\dfrac{\sqrt 3AB^2}{4}=4\sqrt 3\\↔AB^2=16\\↔AB=4(AB>0;cm)\\→MN=2\\→B\)