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7 tháng 5 2023

1.wait

2.would miss

3.would have helped

4.had used

5.would be

6.will play

7.grew

8.would see

9.didn't have to do

10.had stayed up

ʂεμɭ❦

31 tháng 3 2022

\(a;\left(\cos a-\sin a\right)\left(cosa+sina\right)=cos^2a-sin^2a=1-sin^2a-sin^2a=1-2sin^2a\)

\(b;VP=\left(2cosa-1\right)\left(2cosa+1\right)=4cos^2a-1=4\left(1-sin^2a\right)-1=3-4sin^2a=VT\)

e;\(\dfrac{1}{1+tana}+\dfrac{1}{1+cota}=1\Leftrightarrow cota+tana+2=\left(cota+1\right)\left(tana+1\right)\Leftrightarrow cota+tana+2=cota.tana+cota+tana+1\Leftrightarrow cota+tana+2=1+cota+tana+1\Leftrightarrow0=0\left(đúng\right)\Rightarrow VT=VP\)

\(d;sin^3a+cos^3a=\left(sina+cosa\right)\left(sin^2a-sina.cosa+cos^2a\right)=\left(sina+cosa\right)\left(1-sina.cosa\right)\left(đpcm\right)\left(hđt:a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right)\)

\(c;sin^2a.cosa+sina.cos^2a=\left(sina.cosa\right)\left(sin^2+cos^2\right)=sina.cosa\)

\(f;;tana+\dfrac{cosa}{1+sina}=\dfrac{sina}{cosa}+\dfrac{cosa}{1+sina}=\dfrac{sina+sin^2a+cos^2a}{cosa\left(1+sina\right)}=\dfrac{1+sina}{cosa\left(1+sina\right)}=\dfrac{1}{cosa}\)

\(g;1+cot^2a=\dfrac{1}{sin^2a}=\dfrac{1}{1-cos^2a}=\dfrac{1}{\left(1-cosa\right)\left(1+cosa\right)}\left(đpcm\right)\)

\(h;\dfrac{1+cosa}{1-cosa}-\dfrac{1-cosa}{1+cosa}=\dfrac{\left(cosa+1\right)^2-\left(cosa-1\right)^2}{1-cosa^2}=\dfrac{\left(cosa+1-cosa+1\right)\left(cosa+1+cosa-1\right)}{1-cos^2a}=\dfrac{4cosa}{sin^2a}\left(đpcm\right)\)

\(k;\dfrac{1+cosa}{sina}-\dfrac{sina}{1+cosa}=\dfrac{\left(cosa+1\right)^2-sin^2a}{sina\left(1+cosa\right)}=\dfrac{cos^2a+2cosa+1-sin^2a}{sina\left(1+cosa\right)}=\dfrac{2cos^2a+2cosa}{sina\left(1+cosa\right)}=\dfrac{2cosa\left(1+cosa\right)}{sina\left(1+cosa\right)}=\dfrac{2cosa}{sina}=2cota\left(đpcm\right)\)

\(m;;;\Leftrightarrow sin^3a=cosa\left(1+cosa\right)\left(tana-sina\right)=\left(cosa+cos^2a\right)\left(tana-sina\right)\Leftrightarrow sin^3a=\left(cosa+cos^2a\right)\left(\dfrac{sina}{cosa}-sina\right)=sina-sina.cosa+cosa.sina-cos^2a.sina\Leftrightarrow sin^3a=sina-cos^2a.sina\Leftrightarrow sin^3a-sina\left(1-cos^2a\right)=0\Leftrightarrow sin^3a-sina.sin^2a=0\Leftrightarrow0=0\left(đúng\right)\Rightarrowđpcm\)

2 tháng 1 2022

Chữ khó nhìn quá bạn!

1, I suggest collecting boys

2, How about going to the beach

3, Because she was very tired, she went to bed

2 tháng 1 2022

Khoa should use more efficient build

stopping using plastic bags

well

sings very sweetly

dances wonderfully

9 tháng 3 2022

Giúp mk vớiiiiiii

9 tháng 3 2022

18-(50-x)=-23x2

18-(50-x)=-46

50-x=-46+18

50-x=-28

x=50-(-28)

  =50+28

X=78

17 tháng 2 2022

Tham Khảo:

Every year, on January 13, Lim festival is held in Tien Du, Bac Ninh. While the festival is going on, there's a lot of activity. Like other festivals, the Lim festival is divided into ceremonies and festivals. The ceremony is organized with traditional rituals such as worshiping and rituals.

17 tháng 2 2022

refer

Every year, on January 13, Lim festival is held in Tien Du, Bac Ninh. While the festival is going on, there's a lot of activity. Like other festivals, the Lim festival is divided into ceremonies and festivals. The ceremony is organized with traditional rituals such as worshiping and rituals.

19 tháng 2 2020

Ta có\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\)

\(\Leftrightarrow\frac{x-2}{2016}+1+\frac{x-3}{2017}+1+\frac{x-4}{2018}=0\)

\(\Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\) Vì \(\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)>0\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

Mrs. Mai teaches well.
Mai cycles carefully.

26 tháng 12 2021

3.  They asked him " Do you to play football? "

They asked him if he played football.

4. She asked me " Why do you have to do this work ?"

She asked me why I had to do that work.