1. illegal

2. traffic jam

3. seatbelt

4. safely

5. railway station

6. safety

7. traffic signs

8. helicopter

9. tricycle

10. boat

Ai còn thức giúp em với ạ , em sẽ vào trang cá nhân của 2 người đầu tiên và tick 5 lần cho câu trả lời của người đó ạ , người thứ 3 em sẽ tick 2 lần ạ . M.n giúp em với , gấp lắm ạ !

Đến sáng ngày mai ai giúp em thì ko kịp nộp bài nữa nên 12h tối nay và sáng mai em ko tick ạ . Mong m.n thoong cảm và giúp cho em với ạ 

Hết thời gian rồi ạ , pp m.n 

13:

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}sin\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{2pi}{33}\right)\cdot cos\left(\dfrac{4pi}{33}\right)\cdot cos\left(\dfrac{8pi}{33}\right)\cdot cos\left(\dfrac{16pi}{33}\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{4}\cdot sin\dfrac{4}{33}pi\cdot cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{8}\cdot sin\dfrac{8}{33}pi\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{16}\cdot sin\dfrac{16}{33}pi\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{3}\right)}\cdot\dfrac{1}{32}\cdot sin\dfrac{32}{33}pi\)

=1/32

10:

\(=\dfrac{1}{2}\left[cos100+cos60\right]+\dfrac{1}{2}\cdot\left[cos100+cos20\right]\)

=cos100+1/2*cos20+1/4

6:

sin6*cos12*cos24*cos48

=1/cos6*cos6*sin6*cos12*cos24*cos48

=1/cos6*1/2*sin12*cos12*cos24*cos48
=1/cos6*1/4*sin24*cos24*cos48

=1/cos6*1/8*sin48*cos48

=1/cos6*1/16*sin96

=1/16

Bài 3 :

\(\Leftrightarrow\sqrt{9x^2-6x+1}=\sqrt{\left(3x-1\right)^2}=\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

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

Vậy ...

Bài 5 :

Ta có :\(x-5\sqrt{x}+7=x-2.\sqrt{x}.\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

Thấy : \(\left(\sqrt{x}-\dfrac{5}{2}\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow P=\dfrac{1}{x-5\sqrt{x}+7}=\dfrac{1}{\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

Vậy \(Max_P=\dfrac{4}{3}\Leftrightarrow\sqrt{x}-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{25}{4}\)
 

Bài 1: 

a) Ta có: \(\sqrt{25}\cdot\sqrt{144}+\sqrt[3]{-27}-\sqrt[3]{216}\)

\(=5\cdot12-3-6\)

\(=60-9=51\)

b) Ta có: \(\sqrt{8.1\cdot360}\)

\(=\sqrt{8.1\cdot10\cdot36}\)

\(=\sqrt{81\cdot36}\)

\(=9\cdot6=54\)

Bài 2: 

a) Ta có: \(\sqrt{80}-\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3\dfrac{1}{5}}\)

\(=4\sqrt{5}-\sqrt{5}+2+\dfrac{4}{\sqrt{5}}\)

\(=3\sqrt{5}+2+\dfrac{4\sqrt{5}}{5}\)

\(=\dfrac{10+19\sqrt{5}}{5}\)

b) Ta có: \(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}\)

\(=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13\left(4-\sqrt{3}\right)}{\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)}\)

\(=-\sqrt{3}+\sqrt{3}+6-4+\sqrt{3}\)

\(=2+\sqrt{3}\)

Bài 3: 

Ta có: \(\sqrt{9x^2+6x+1}=5\)

\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=5\)

\(\Leftrightarrow\left|3x+1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=5\\3x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-2\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{4}{3};-2\right\}\)

Bài 4: 

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{4\sqrt{x}-8}{1-x}\)

\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{4\sqrt{x}-8}{-\left(x-1\right)}\)

\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4\sqrt{x}}{4\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

câu 5: 

x=3,6

y=6,4

câu 6: chụp lại đề

câu 7:

a)ĐKXĐ: \(x\ge0\)

\(3\sqrt{x}=\sqrt{12}\\ \Rightarrow9x=12\\ \Rightarrow x=\dfrac{4}{3}\)

b) ĐKXĐ: \(x\ge6\)

\(\sqrt{x-6}=3\\ \Rightarrow x-6=9\\ \Rightarrow x=15\)

Câu 5: 

Áp dụng định lý Pi-ta-go ta có:

\(AB^2+AC^2=BC^2\\ \Rightarrow BC=\sqrt{6^2+8^2}\\ \Rightarrow BC=10\)

Áp dụng HTL ta có: \(x.BC=AB^2\Rightarrow x.10=6^2\Rightarrow x=3,6\)

Áp dụng HTL ta có: \(x.BC=AC^2\Rightarrow x.10=8^2\Rightarrow x=6,4\)