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III. Correct a mistake1. I bought a ticket for that film this morning. I will see it tonight.2. Don't talk loudly. My daughter was studying in the next room.3. Roles in homemaking and breadwinning can share by family members.4. It didn’t spend her much time to do the laundry because she has a washing machine.5. I shall contact you as soon as they will phone me.6. Our new neighbour is a bored man; he keeps talking all the time.7. In space, with no gravity for muscles to work against, the body...
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III. Correct a mistake

1. I bought a ticket for that film this morning. I will see it tonight.

2. Don't talk loudly. My daughter was studying in the next room.

3. Roles in homemaking and breadwinning can share by family members.

4. It didn’t spend her much time to do the laundry because she has a washing machine.

5. I shall contact you as soon as they will phone me.

6. Our new neighbour is a bored man; he keeps talking all the time.

7. In space, with no gravity for muscles to work against, the body becomes weakly.

8. Animism is the belief that objects and natural phenomena such as rivers, rocks, and wind are live and have feelings.

9. High blood pressure results from either an increased output of blood from the heart and an increased resistance to its flow through tiny branches of the arteries.

10. Her passion for helping people have motivated her to found her own charity organization.

0
28 tháng 7 2021

\(mx^2-2\left(m+2\right)x+2m-1< 0\)

\(< =>mx^2-2\left(m+2\right)x+2m-1\ge0\)

\(a=m\ne0\)

\(\Delta=\left(2m+2\right)^2-4m\left(2m-1\right)\)

\(\Delta=4m^2+8m+4-8m^2+4m\)

\(\Delta=12m-4m^2+4\)

\(< =>\hept{\begin{cases}a>0\\\Delta\le0\end{cases}\hept{\begin{cases}m>0\\12m-4m^2+4\le0\end{cases}\hept{\begin{cases}m>0\\m=\left[\frac{3-\sqrt{13}}{2};\frac{3+\sqrt{13}}{2}\right]\end{cases}}}}\)

\(< =>m=(0;\frac{3+\sqrt{13}}{2}]\)

vậy m vô số nghiệm để bpt vô nghiệm

I C A H J B

Ta có:

 \(AJ=\frac{1}{2}\left(AH+AC\right)\)

\(BI=\frac{1}{2}\left(BA+BH\right)\)

\(AJ.BI=\frac{1}{4}\left(AH+AC\right)\left(BA+BH\right)\)

\(=\frac{1}{4}\left(AH.BA+AH.BH+AC.BA+AC.BH\right)\)

\(=\frac{1}{4}\left(AH.HA+0+0+HC.BH\right)\)

\(=\frac{1}{4}\left(-AH^2+BH.HC\right)\)

Mà: Tam giác ABC vuông tại A <=> \(AH^2=HC.HB\)

Vậy \(AJ.BI=0\Leftrightarrow AJ\perp BI\)