i: ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}+1\\x\le-\sqrt{5}+1\end{matrix}\right.\)

l:ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-5\end{matrix}\right.\)

m: ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le3\end{matrix}\right.\)

n: ĐKXĐ: \(x\in R\)

1 eats => has eaten

2 buy => bought

3 took => take

4 ate => eat

5 and => so

6 so => as

7 photo => photography 

8 are, either => were, too

9 the puppets => puppets

10 take => takes

d: \(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)

=>x+100=0

hay x=-100

a: \(\Leftrightarrow2\left(x+1\right)^2+3\left(x^2-3x+2\right)=\left(3x-1\right)\left(x-4\right)+56\)

\(\Leftrightarrow2x^2+4x+2+3x^2-9x+6-\left(3x^2-13x+4\right)-56=0\)

\(\Leftrightarrow5x^2-5x-48-3x^2+13x-4=0\)

\(\Leftrightarrow2x^2+8x-52=0\)

\(\Leftrightarrow x^2+4x-13=0\)

\(\Leftrightarrow\left(x+2\right)^2=17\)

hay \(x\in\left\{-\sqrt{17}-2;\sqrt{17}-2\right\}\)

Pt hoành độ giao điểm:

\(x^2=-2\left(m-2\right)x-m^2+4m\Leftrightarrow x^2+2\left(m-2\right)x+m^2-4m=0\) (1)

\(\Delta'=\left(m-2\right)^2-\left(m^2-4m\right)=4>0;\forall m\Rightarrow\) (1) luôn có 2 nghiệm pb hay (d) luôn cắt (P) tại 2 điểm pb

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-2\right)\\x_1x_2=m^2-4m\end{matrix}\right.\)

Để biểu thức đề bài xác định \(\Rightarrow x_1x_2\ne0\Rightarrow\left[{}\begin{matrix}m\ne0\\m\ne4\end{matrix}\right.\)

Khi đó:

\(\dfrac{3}{x_1}+x_2=\dfrac{3}{x_2}+x_1\Leftrightarrow\left(3+x_1x_2\right)x_2=\left(3+x_1x_2\right)x_1\)

\(\Leftrightarrow\left(3+x_1x_2\right)\left(x_1-x_2\right)=0\)

\(\Leftrightarrow3+x_1x_2=0\) (do \(\Delta>0\) nên \(x_1-x_2\ne0\) với mọi m)

\(\Leftrightarrow3+m^2-4m=0\Rightarrow\left[{}\begin{matrix}m=1\\m=3\end{matrix}\right.\)