câu d tìm ra x,y là bao nhiêu

a: \(\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\3x+12y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=2-\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)

8:

\(=\dfrac{cos10-\sqrt{3}\cdot sin10}{sin10\cdot cos10}=\dfrac{2\left(\dfrac{1}{2}\cdot cos10-\dfrac{\sqrt{3}}{2}\cdot sin10\right)}{sin20}=\dfrac{sin\left(30-10\right)}{sin20}=1\)

10:

\(=\left(2-\sqrt{3}\right)^2+\left(2+\sqrt{3}\right)^2\)

=7-4căn 3+7+4căn 3=14

12:

\(=cos^270^0+\dfrac{1}{2}\left[cos60-cos140\right]\)

\(=cos^270^0+\dfrac{1}{2}\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot2cos^270^0+\dfrac{1}{.2}\)

=1/4+1/2=3/4

1. illegal

2. traffic jam

3. seatbelt

4. safely

5. railway station

6. safety

7. traffic signs

8. helicopter

9. tricycle

10. boat

bạn ơi bài năm phần nào vậy 

1.an

2.a

3.the

4.the

5.the

6.the

ko phải bn ơi ý là bài đó hoàn thành rồi h đặt câu hỏi cho từng câu trong bài á

Bài 1:

a. \(=2\sqrt{3^2}+\sqrt{15}-\sqrt{4.15}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)

b. \(=5\sqrt{10}+2\sqrt{5^2}-\sqrt{25.10}=5\sqrt{10}+10-5\sqrt{10}=10\)

c. \(=\left(\sqrt{4.7}-\sqrt{4.3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=2\sqrt{7^2}-2\sqrt{21}-\sqrt{7^2}+2\sqrt{21}=7\)

d. \(=\left(\sqrt{9.11}-\sqrt{9.2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=3\sqrt{11^2}-3\sqrt{22}-\sqrt{11^2}+3\sqrt{22}=22\)

Bài 3:

a.

\(=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+\sqrt{x}^2\right)=1-\sqrt{x}^3=1-x\sqrt{x}\)

b.

\(=\left(\sqrt{x}+2\right)\left(\sqrt{x}^2-2\sqrt{x}+2^2\right)=\sqrt{x}^3+2^3=x\sqrt{x}+8\)

c.

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}^2+\sqrt{xy}+\sqrt{y}^2\right)=x\sqrt{x}-y\sqrt{y}\)

d.

\(=\left(x+\sqrt{y}\right)\left(x^2-x\sqrt{y}+\sqrt{y}^2\right)=x^3+y\sqrt{y}\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Bài 6: 

a: Xét tứ giác DEBF có 

DE//BF

DE=BF

Do đó: DEBF là hình bình hành

Bài 5 Mn ơi