a, \(2n+5⋮n-1\)

\(2\left(n-1\right)+7⋮n-1\)

\(7⋮n-1\)hay \(n-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

b, Công thức tổng quát : \(A\left(x\right).B\left(x\right)=0\Rightarrow\orbr{\begin{cases}A\left(x\right)=0\\B\left(x\right)=0\end{cases}}\)

\(\left(2n+3\right)\left(n-4\right)=0\Leftrightarrow\orbr{\begin{cases}n=-\frac{3}{2}\\n=4\end{cases}}\)

c, \(\left|x-3\right|< 3\Leftrightarrow-3< x-3< 3\)

\(\Leftrightarrow-3+3< x< 3+3\Leftrightarrow0< x< 6\)

Vậy \(x\in\left\{1;2;3;4;5;\right\}\)

giải chi tiết ra giúp mk nhé các bn!thanks các bn nhiều ^^

câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

          \(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);

          \(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\) 

         \(\dfrac{1}{x^2-1}\)  =  \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

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

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

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

chữ xấu quá khum đọc đc

bé đến lớn hay lớn đến bé

\(a.\left|x-2\right|+3=x.\\ \Leftrightarrow\left|x-2\right|=x-3.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3>0.\\\left(\left|x-2\right|\right)^2=\left(x-3\right)^2.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x^2-4x+4=x^2-6x+9.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\2x=5.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x=\dfrac{5}{2}.\end{matrix}\right.\) \(\Leftrightarrow x\in\phi.\)

\(b.\left(3x-4\right)\left(2x-5\right)=\left(3x-4\right)\left(x+2\right).\\ \Leftrightarrow\left(3x-4\right)\left(2x-5-x-2\right)=0.\\ \Leftrightarrow\left(3x-4\right)\left(x-7\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}.\\x=7.\end{matrix}\right.\)

\(\dfrac{x}{x-2}+\dfrac{x-1}{x}=2.\left(x\ne2;0\right).\\ \Leftrightarrow\dfrac{x^2+\left(x-1\right)\left(x-2\right)-2x\left(x-2\right)}{x\left(x-2\right)}=0.\\ \Rightarrow x^2+x^2-2x-x+2-2x^2+4x=0.\\ \Leftrightarrow x=-2\left(TM\right).\)

\(d.\dfrac{x-2}{2}-\dfrac{x+5}{3}=1-\dfrac{x-2}{4}.\\ \Leftrightarrow\dfrac{6x-12-4x-20-12+3x-6}{12}=0.\\ \Rightarrow5x=50.\\ \Leftrightarrow x=10.\)

9:

a: XétΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

=>BA/BH=BC/BA
=>BA^2=BH*BC

b: BC=25cm; AB=căn 9*25=15cm; AC=căn 16*25=20cm

S ABC=1/2*15*20=150cm²

C ABC=25+15+20=60cm

Câu 36 C

Câu 37 C

Câu 38 C

11. Have you ever been 

12. haven't done

13. have you seen - has already done

14. have just decided

15. has been

16. hasn't had

17. hasn't played

18. haven't had

19. haven't seen

20. have just realized

Đây là thì HTHT nhé, cấu trúc rất dễ nhớ thôi nè :3 

      S+ have/has + V_p2 (nói về một việc đã bắt đầu trong quá khứ và vẫn tiếp diễn đến bh)

Dấu hiệu nhận biết: for + một khoảng thời gian, since + một thời gian cụ thể trong quá khứ

vd: She hasn't played badminton for 2 years.

 He hasn't gone to school since 3 weeks ago.

he would join the Science Club the day after
his room is untidy, his mother is unhappy
started our work, she had explained everything clearly
I had enough money, I could buy this motorbike