giúp e vs m.n ơi 
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
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)
\(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\) đk \(x\) ≠ 2; -2
\(\dfrac{x}{2x-4}\) = \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\)
\(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)
\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)
9 tháng 3 2022
\(a.\left|x-2\right|+3=x.\\ \Leftrightarrow\left|x-2\right|=x-3.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3>0.\\\left(\left|x-2\right|\right)^2=\left(x-3\right)^2.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x^2-4x+4=x^2-6x+9.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\2x=5.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x=\dfrac{5}{2}.\end{matrix}\right.\) \(\Leftrightarrow x\in\phi.\)
\(b.\left(3x-4\right)\left(2x-5\right)=\left(3x-4\right)\left(x+2\right).\\ \Leftrightarrow\left(3x-4\right)\left(2x-5-x-2\right)=0.\\ \Leftrightarrow\left(3x-4\right)\left(x-7\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}.\\x=7.\end{matrix}\right.\)
\(\dfrac{x}{x-2}+\dfrac{x-1}{x}=2.\left(x\ne2;0\right).\\ \Leftrightarrow\dfrac{x^2+\left(x-1\right)\left(x-2\right)-2x\left(x-2\right)}{x\left(x-2\right)}=0.\\ \Rightarrow x^2+x^2-2x-x+2-2x^2+4x=0.\\ \Leftrightarrow x=-2\left(TM\right).\)
\(d.\dfrac{x-2}{2}-\dfrac{x+5}{3}=1-\dfrac{x-2}{4}.\\ \Leftrightarrow\dfrac{6x-12-4x-20-12+3x-6}{12}=0.\\ \Rightarrow5x=50.\\ \Leftrightarrow x=10.\)
17 tháng 6 2023
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
H
18 tháng 10 2021
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 + Vp2 (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.
MK
20 tháng 10 2021
1.old 2.themselves 3.enough 4.see 5.on 6. long black 7.of 8.get 9.herself 10.did 11.suitable 12.to wear 13.on
S
_silverlining
CTVVIP
12 tháng 5 2022
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
\(a,=\left(x-y\right)\left(x+y\right)+7\left(x-y\right)=\left(x+y+7\right)\left(x-y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\)