1/
a/ $AD$ là đường phân giác $\widehat A$
$→\dfrac{AB}{AC}=\dfrac{DB}{DC}$ hay $\dfrac{4}{6}=\dfrac{DB}{DC}$
$→\dfrac{2}{3}=\dfrac{DB}{DC}\\↔\dfrac{DB}{2}=\dfrac{DC}{3}$
Áp dụng tính chất dãy tỉ số bằng nhau
$\dfrac{DB}{2}=\dfrac{DC}{3}=\dfrac{DB+DC}{2+3}=\dfrac{BC}{5}=\dfrac 5 5=1$
$→DC=3cm$
b/ $DE//AB$
$→\dfrac{DC}{BC}=\dfrac{DE}{AB}$ hay $\dfrac{3}{5}=\dfrac{DE}{4}$
$↔DE=\dfrac{4.3}{5}=\dfrac{12}{5}cm$
c/ $DE//AB$
$→ΔEDC\backsim ΔABC$
$→\dfrac{S_{ΔEDC}}{S_{ΔABC}}=\bigg(\dfrac{DB}{BC}\bigg)^2=\bigg(\dfrac{3}{5}\bigg)^2=\dfrac{9}{25}$
Vậy $\dfrac{S_{ΔEDC}}{S_{ΔABC}}=\dfrac{9}{25}$
2/ ĐK: $x\ge 0,x\ne 4$
$A=\dfrac{\sqrt x}{\sqrt x -2}\\=\dfrac{\sqrt x -2+2}{\sqrt x -2}\\=1+\dfrac{2}{\sqrt x -2}$
$A∈\Bbb Z\\→1+\dfrac{2}{\sqrt x -2}∈\Bbb Z\\→\dfrac{2}{\sqrt x -2}∈\Bbb Z\\→2\vdots \sqrt x -2\\→\sqrt x -2∈Ư(2)=\{±1;±2\}$
Ta có bảng:
$\begin{array}{|c|c|c|}\hline \sqrt x -2&1&-1&2&-2\\\hline \sqrt x&3&1&4&0\\\hline x&9&1&16&0\\\hline\end{array}$
mà $x\ge 0,x\ne 4$
$→x∈\{9;1;16;0\}$
Vậy $x∈\{9;1;16;0\}$ thì $A∈\Bbb Z$
