A=\(\dfrac{3x+2}{x-3}=\dfrac{3.\left(x-3\right)+11}{x-3}=3+\dfrac{11}{x-3}\)(x\(\ne3\))

để A nguyên<=> x-3 \(\in\)Ư₍₁₁₎={\(\pm11,\pm1\)}

=>\(\left[{}\begin{matrix}x-3=11\\x-3=-11\\x-3=1\\x-3=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=14\left(tm\right)\\x=-8\left(tm\right)\\x=4\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

vậy x\(\in\left\{14,4,2,-8\right\}\)thì A nguyên

ta có: tan B=\(\dfrac{8}{15}\)

=>tan B=\(\dfrac{8}{15}=\dfrac{AC}{AB}\)

mà AB=30 cm (gt)

=> AC= 8.30:15=16 cm

xét tam giác ABC vuông tại A (gt) 

=> AC²+AB²=BC²  ( Định lí pytago)

hay 16²+30²=BC²

=>BC=\(\sqrt{16^2+30^2}=34\)

ta có sin B=\(\dfrac{AC}{CB}=\dfrac{16}{34}=\dfrac{8}{17}\)

cos B= \(\dfrac{AB}{BC}=\dfrac{30}{34}=\dfrac{15}{17}\)

cotg B =\(\dfrac{30}{16}=\dfrac{15}{8}\)

''tôi cũng vậy'' trong câu phủ định:

 either + do/ does /did +S ( do/does/ did phụ thuộc vào S)

14.

=\(\dfrac{-\left(7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{-7x-21\sqrt{x}-14}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{10x-12\sqrt{x}+2}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+​​​​​​\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)​​​​​​​​

=\(\dfrac{-7x-21\sqrt{x}-14+10x-12\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{3x-6\sqrt{x}}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)=\(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

17.

16. \(\dfrac{2\sqrt{x}-4}{3\sqrt{x}-4}-\dfrac{4+2\sqrt{x}}{\sqrt{x}-2}+\dfrac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

=\(\dfrac{\left(2\sqrt{x}-4\right)\left(\sqrt{x}-2\right)-\left(4+2\sqrt{x}\right)\left(3\sqrt{x}-4\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=\(\dfrac{2x-8\sqrt{x}+8-\left(4\sqrt{x}+6x-16\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=​\(\dfrac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)​​​

=\(\dfrac{-3x+\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(x+2\right)}\)

=\(\dfrac{-\left(3x+3\sqrt{x}-4\sqrt{x}-4\right)}{\left(3\sqrt{x}-4\right)\left(x+2\right)}\)

=\(\dfrac{-\left(3\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}+2\right)}\)=\(\dfrac{-\sqrt{x}-1}{\sqrt{x}+2}\)

biện pháp : nhân hóa , điệp ngữ

từ láy:

phanh phách, hủn hoẳn, phành phạch,rung rinh, ngoàm ngoạp, giòn giã, bè bè 

biện pháp tu từ :nhân hóa.

chonj B

\(\dfrac{24}{16},\dfrac{11}{8},\dfrac{5}{8},\dfrac{2}{4}\)