giúp em vs ạ T^T
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Dùng Ca(OH)2 em nha vì khí CO2 và SO2 đều tác dụng được với dd Ca(OH)2
2 She said nothing grew in her garden because it never got any sun
3 He told his mother he was going to HN the day after
6 He told her to switch off the TV
9 I warned them not to watch late night horror film
Tham khảo: Trả lời: – Rừng có độ dốc lớn hơn 15 độ, nơi rừng phòng hộ không được khai thác trắng vì gây ra xói mòn, rửa trôi. – Khai thác rừng nhưng không trồng rừng ngay thì sẽ làm cho đất bị thoái hóa, rữa trôi, xói mòn, có thể gây ra lũ lụt,….
Câu 2:
\(a,\Leftrightarrow\Delta'=\left(1-m\right)^2-\left(m^2-m\right)>0\\ \Leftrightarrow m^2-2m+1-m^2+m>0\\ \Leftrightarrow1-m>0\Leftrightarrow m< 1\\ b,\text{Áp dụng Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(1-m\right)\\x_1x_2=m^2-m\end{matrix}\right.\\ \left(2x_1-1\right)\left(2x_2-1\right)-x_1x_2=1\\ \Leftrightarrow2x_1x_2-2\left(x_1+x_2\right)+1-x_1x_2=1\\ \Leftrightarrow x_1x_2-2\left(x_1+x_2\right)=0\\ \Leftrightarrow m^2-m-4\left(1-m\right)=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow\left(m-1\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-4\left(tm\right)\end{matrix}\right.\)
Vậy m=-4
Câu 1:
\(1,\Leftrightarrow2x-2=3\Leftrightarrow x=\dfrac{5}{2}\\ 2,ĐK:x\ne\pm1\\ PT\Leftrightarrow\dfrac{2x^2+2x-1}{x^2-1}=2\\ \Leftrightarrow2x^2+2x-1=2x^2-2\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\\ 3,\Leftrightarrow\left[{}\begin{matrix}3x-2=2x-1\\3x-2=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)
\(4,\Leftrightarrow\left[{}\begin{matrix}3x-1=2-x\left(x\ge\dfrac{1}{3}\right)\\3x-1=x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(tm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\\ 5,\Leftrightarrow4x^2-2x+10=9x^2-6x+1\left(x\le\dfrac{1}{3}\right)\\ \Leftrightarrow5x^2-4x-9=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(6,\Leftrightarrow3x^2-9x+1=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow2x^2-5x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ 7,\Leftrightarrow2x^2+3x-4=7x+2\left(x\ge-\dfrac{2}{7}\right)\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
refer
Rosa Hi, what are you doing this evening? do
..........Do....... I......come...... (1) round? come
Maria Not this evening. I .....am........ (2) busy till late be
Rosa When do you think it......is...... (3) convenient for me to pop round? be
Maria Well,......do...... we .........check........ (4) the date? check
Have you got your diary handy? Now, let’s see. Today is
Tuesday the 20th so tomorrow .....will be.............. (5) Wednesday 21st. be
I ....am...... (6) so busy tomorrow – what about you? Do you think be
you .......are...... (7) free? be
Rosa I ...........will see........... (8) my dentist tomorrow. Is Thursday OK? see
Maria Yeah. I think that ........is........(9) fine. be
Rosa OK. What time ....do......... I .......come................. (10) round? come
Maria I .......leaves........ (11) the house at all on Thursday so I don’t think leave
it ....is....... (12) a problem, whatever time you come. be
Rosa That’s fine.
Maria And ......do..... you .........bring......... (13) the manuscript with you? bring
Rosa Don’t worry. I ......am forgetting....... (14) it. forget
Maria OK. I .........see.......... (15) you Thursday then. Cheers. see
1 Shall I come
2 will be
3 will be
4 shall we check
5 will be
6 am going to by - will be
8 am going to see
9 will be
9 shall I come
11 will be - is
13 Will you bring
14 win't forget
15 will see
a) 4,7 x 5,5 - 4,7 x 4,5=4,7 x (5,5 - 4,5) = 4,7 x1 = 4,7
b) 7,8 x 0,35 + 0,35 x 2,2 = 0,35 x (7,8 + 2,2) = 0,35 x 10 = 3,5
a) 4,7 x 5,5 - 4,7 x 4,5
= 4,7 x (5,5 - 4,5)
= 4,7 x 1
= 4,7
b) 7,8 x 0,35 + 0,35 x 2,2
= 0,35 x (7,8 + 2,2)
= 0,35 x 10
= 3,5
a/ Xét \(\Delta HAC\) và \(\Delta ABC\) có
\(\widehat{BAH}=\widehat{ACH}\) (Vì cùng phụ với \(\widehat{HAC}\) ) => \(\Delta BAH\) đồng dạng với \(\Delta ABC\)
\(\Rightarrow\frac{AH}{AB}=\frac{AC}{BC}\Rightarrow AH.BC=AB.AC\left(dpcm\right)\)
b/ Ta có
\(HK=CK;HI=AI\) => KI là đường trung bìcuarHHAC tg HAC => KI//AC\(\Rightarrow\widehat{HKI}=\widehat{BCA}\)
Xét tg vuông HKI và tg vuông ABC có
\(\widehat{HKI}=\widehat{BAC}\left(cmt\right)\) => tg HKI đồng dạng với tg ABC
Bài nào