1c  2.a  3b  4a

Còn bài II nữa ạ , bạn giúp mình nốt nhá 

:v bạn tách nhỏ đề ra và chụp ngang ra nhé

Okee 👌

1: \(\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)

a: Xét ΔMAB và ΔMDC có

\(\hat{MAB}=\hat{MDC}\) (hai góc đồng vị, AB//DC)

\(\hat{AMB}\) chung

Do đó: ΔMAB~ΔMDC

=>\(k=\frac{AB}{CD}=\frac{3}{6.5}=\frac{6}{13}\)

b: ΔMAB~ΔMDC

=>\(\frac{MA}{MD}=\frac{AB}{DC}\)

=>\(MA\cdot DC=MD\cdot AB\)

c: Khoảng cách giữa hai đáy của hình thang ABCD là:

\(h=S_{ABCD}\cdot\frac{2}{AB+CD}=19\cdot\frac{2}{3+6,5}=19\cdot\frac{2}{9,5}=2\cdot2=4\left(\operatorname{cm}\right)\)

ΔMAB~ΔMDC

=>\(\frac{S_{MAB}}{S_{MDC}}=\left(\frac{AB}{DC}\right)^2=\left(\frac{3}{6.5}\right)^2=\left(\frac{6}{13}\right)^2=\frac{36}{169}\)

=>\(\frac{S_{MAB}}{S_{ABCD}}=\frac{36}{169-36}=\frac{36}{133}\)

=>\(\frac{S_{AMB}}{19}=\frac{36}{133}\)

=>\(S_{AMB}=36\times\frac{19}{133}=\frac{36}{7}\left(\operatorname{cm}^2\right)\)

Đề bài là: Tính cos2x 

Cảm ơn mn nhiều ạ!

`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`

`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`

`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`

`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`

`<=> cos2x - cos4x - cos2x = 1/2`

`<=> cos4x = cos(2π)/3`

`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)

1 There are 5000 living languages in the world

2 It is Chinese

3 It is English

4 Yes, because I think English is interesting

V

1 He used to play the guitar at night

2 She wishes she had a pen pal

3 We started learning E 4 years ago

4 How long have you had that car

c)\(\left\{{}\begin{matrix}u_1+u_3=3\\u_1^2+u_3^2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_3=3\\\left(u_1+u_3\right)^2-2u_1u_3=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_3=3\\u_1u_3=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}u_1=2\\u_3=1\end{matrix}\right.\\\left\{{}\begin{matrix}u_1=1\\u_3=2\end{matrix}\right.\end{matrix}\right.\)

Làm nốt (sử dụng công thức: \(u_n=u_1+\left(n-1\right)d\) để tìm được công sai

\(S_n=nu_1+\dfrac{n\left(n-1\right)}{2}d\) để tính tổng 15 số hạng đầu)

d)\(\left\{{}\begin{matrix}u_1+u_2+u_3=14\\u_1u_2u_3=64\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_2-d+u_2+u_2+d=14\\\left(u_2-d\right)u_2\left(u_2+d\right)=64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_2=\dfrac{14}{3}\\\left(u_2^2-d^2\right)u_2=64\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\dfrac{14}{3}=u_2=u_1+d\\d=\dfrac{2\sqrt{889}}{21}\end{matrix}\right.\\\left\{{}\begin{matrix}\dfrac{14}{3}=u_1+d\\d=\dfrac{-2\sqrt{889}}{21}\end{matrix}\right.\end{matrix}\right.\) 

(Làm nốt,số xấu quá)

e)\(\left\{{}\begin{matrix}u_1+u_2+u_3=7\\u_1^2+u_2^2+u_3^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_2+u_3=7\\u_1u_2u_3=\dfrac{21-\left(u_1+u_2+u_3\right)^2}{2}=-14\end{matrix}\right.\)

Làm như ý d)

bạn tính thử cho mình tổng câu c được không ạ, tại mình chưa hiểu lắm á