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

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

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

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

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

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

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

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

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

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

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

Khoa should use more efficient build

stopping using plastic bags

well

sings very sweetly

dances wonderfully

Giúp mk vớiiiiiii

18-(50-x)=-23x2

18-(50-x)=-46

50-x=-46+18

50-x=-28

x=50-(-28)

  =50+28

X=78

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.

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.

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.

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.

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.