16, a

17, a

18, d

19, a

20, b

21, c

17 B: (động vật hoang dã)

19 A:(khí hậu mưa)

20 B(bị động và expect+to-v)

21 C(động lực)

16 A

17 A

17 D

19 A

20 B

21 C

working

directly

Working ( stop+V-ing)
Directly

1) ĐKXĐ: \(x\ge0\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge4\)

4) ĐKXĐ: \(x>16\)

5) ĐKXĐ: \(\left[{}\begin{matrix}x\le-2\\x\ge0\end{matrix}\right.\)

6) ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge4\end{matrix}\right.\)

7) ĐKXĐ: \(\left[{}\begin{matrix}1\le x\\x< 3\end{matrix}\right.\)

8) ĐKXĐ: \(\left[{}\begin{matrix}x\le-2\\x>3\end{matrix}\right.\)

9) ĐKXĐ: \(x\in R\)

10) ĐKXĐ: \(x\in R\)

11) ĐKXĐ: \(x\in R\)

12) ĐKXĐ: \(x\in R\)

13) ĐKXĐ: \(x\in R\)

14) ĐKXĐ: \(x\in R\)

15) ĐKXĐ: \(x\in R\)

16) ĐKXĐ: \(x\ne-\dfrac{1}{2}\)

17) ĐKXĐ: \(x\ge7\)

18) ĐKXĐ: \(x\ge-5\)

19) ĐKXĐ: \(x>5\)

20) ĐKXĐ: \(x>\dfrac{7}{3}\)

21) ĐKXĐ: \(x\le\dfrac{5}{2}\)

22) ĐKXĐ: \(x\le\dfrac{4}{3}\)

\(1,\Leftrightarrow x^2-8x+16-x^2+x+12=7\\ \Leftrightarrow-7x=-21\\ \Leftrightarrow x=3\\ 2,\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

a) \(\Leftrightarrow x^2+10x+25-x^2+8x-15=-8\\ \Leftrightarrow18x=-18\\ \Leftrightarrow x=-1\)

b) \(\Leftrightarrow\left(2x+1\right)^2-3\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Chụp rõ đề hơn đi bạn.

\(1,\\ a,\dfrac{8x}{2xy}=\dfrac{4x}{y}\\ b,\dfrac{2xy}{6y}=\dfrac{x}{3}\\ c,\dfrac{3\left(x+2\right)}{2x}=\dfrac{6\left(x+2\right)}{4x}\\ d,\dfrac{4\left(x-2\right)}{3\left(x+1\right)}=\dfrac{8\left(x-2\right)x}{6\left(x+1\right)x}\\ 2,\\ \dfrac{x^2+3x+2}{x^2+x}=\dfrac{x^2+x+2x+2}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)}=\dfrac{x+2}{x}\\ 3,\\ \dfrac{x^2-3x}{x^2-9}=\dfrac{x}{x+3}\)

Bài 3: 

Ta có: \(x^2-2x+4=\left(x-1\right)^2+3\ge3\forall x\)

\(\Leftrightarrow P=\dfrac{15}{x^2-2x+4}=\dfrac{15}{\left(x-1\right)^2+3}\le5\forall x\)

Dấu '=' xảy ra khi x=1