11c.

Từ đề bài ta có:

\(\left\{{}\begin{matrix}\dfrac{16a-b^2}{4a}=\dfrac{9}{2}\\16a+4b+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2b^2=-4a\\b=-4a-1\end{matrix}\right.\)

\(\Rightarrow2b^2-b=1\Leftrightarrow2b^2-b-1=0\Rightarrow\left[{}\begin{matrix}b=1\Rightarrow a=-\dfrac{1}{2}\\b=-\dfrac{1}{2}\Rightarrow a=-\dfrac{1}{8}\end{matrix}\right.\)

Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=-\dfrac{1}{2}x^2+x+4\\y=-\dfrac{1}{8}x^2-\dfrac{1}{2}x+4\end{matrix}\right.\)

4f.

Từ đề bài ta có:

\(\left\{{}\begin{matrix}1+b+c=0\\\dfrac{4c-b^2}{4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=-b-1\\c=\dfrac{b^2}{4}-1\end{matrix}\right.\)

\(\Rightarrow\dfrac{b^2}{4}+b=0\)

\(\Rightarrow\left[{}\begin{matrix}b=0\Rightarrow c=-1\\b=-4\Rightarrow c=3\end{matrix}\right.\)

Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=x^2-1\\y=x^2-4x+3\end{matrix}\right.\)

Kẻ AH⊥BC

ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)

\(10,\\ a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\\ \Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge0\\ \Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow\dfrac{a}{2}=b=c=d=e\)

\(4,\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{4}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\ge3ab+3bc+3ca\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{2}a^2-\dfrac{1}{2}b^2-\dfrac{1}{2}c^2-ab-bc-ac\ge0\\ \Leftrightarrow\dfrac{1}{2}a^2+\dfrac{1}{2}b^2+\dfrac{1}{2}c^2+ab+ac+bc\ge0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge0\\ \Leftrightarrow\left(a+b+c\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a+b+c=0\)

\(1,a^2+b^2+c^2\ge ab+bc+ca\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

\(2,a^2+b^2+c^2+3\ge2\left(a+b+c\right)\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\ge0\\ \Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b=c=1\)

\(3,\Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\\ \Leftrightarrow9\left(a^2+b^2+c^2\right)\ge3\left(a+b+c\right)^2\\ \Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\\ \Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\left(luôn.đúng.do.câu.1\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

\(4,\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\ge3a^2bc+3ab^2c+3abc^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2-a^2bc-ab^2c-abc^2\ge0\\ \Leftrightarrow2a^2b^2+2b^2c^2+2a^2c^2-2a^2bc-2ab^2c-2abc^2\ge0\\ \Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(a^2b^2-2ab^2c+b^2c^2\right)+\left(b^2c^2-2abc^2+a^2c^2\right)\ge0\\ \Leftrightarrow\left(ab-ac\right)^2+\left(ab-bc\right)^2+\left(bc-ac\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow ab=bc=ac\)

She does not go to school at 8 o'clock
He does not have dinner at 9 o'clock
She does not go to bed at 7 o'clock
He does not go home at 4 o'clock
He does not go to bed at 9 o'clock

a: |x|=5,6

=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)

c: \(\left|x\right|=3\dfrac{1}{5}\)

=>\(\left|x\right|=3,2\)

=>\(\left[{}\begin{matrix}x=3,2\\x=-3,2\end{matrix}\right.\)

d: |x|=-2,1

mà -2,1<0

nên \(x\in\varnothing\)

d: |x-3,5|=5

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

=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)

e: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=>\(\left|x+\dfrac{3}{4}\right|=\dfrac{1}{2}\)

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

f: \(\left|4x\right|-\left|-13,5\right|=\left|2\dfrac{1}{4}\right|\)

=>\(4\left|x\right|=2,25+13,5=15,75\)

=>\(\left|x\right|=\dfrac{63}{16}\)

=>\(x=\pm\dfrac{63}{16}\)

g: \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\)

=>\(\dfrac{5}{6}-\left|x-2\right|=\dfrac{1}{3}\)

=>\(\left|x-2\right|=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{1}{2}\)

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

h: \(\left|x-\dfrac{2}{5}\right|+\dfrac{1}{2}=\dfrac{3}{4}\)

=>\(\left|x-\dfrac{2}{5}\right|=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{1}{4}\\x-\dfrac{2}{5}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{13}{20}\\x=-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{-5+8}{20}=\dfrac{3}{20}\end{matrix}\right.\)

i: \(\left|5-3x\right|+\dfrac{2}{3}=\dfrac{1}{6}\)

=>\(\left|3x-5\right|=\dfrac{1}{6}-\dfrac{2}{3}=\dfrac{1}{6}-\dfrac{4}{6}=-\dfrac{3}{6}=-\dfrac{1}{2}< 0\)

=>\(x\in\varnothing\)

k: \(-2,5+\left|3x+5\right|=-1,5\)

=>|3x+5|=-1,5+2,5=1

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

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

m: \(\dfrac{1}{5}-\left|\dfrac{1}{5}-x\right|=\dfrac{1}{5}\)

=>\(\left|\dfrac{1}{5}-x\right|=\dfrac{1}{5}-\dfrac{1}{5}=0\)

=>\(\dfrac{1}{5}-x=0\)

=>\(x=\dfrac{1}{5}\)

n: \(-\dfrac{22}{15}x+\dfrac{1}{3}=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\)

=>\(-\dfrac{22}{15}x+\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{5}\)

=>\(-\dfrac{22}{15}x=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)

=>-22x=2

=>\(x=-\dfrac{1}{11}\)

Lời giải:

\(\lim\limits_{x\to 2-}y=\lim\limits_{x\to 2-}\frac{\sqrt{4-x^2}}{(x-2)(x-3)}=\lim\limits_{x\to 2-}\frac{\sqrt{2+x}}{\sqrt{2-x}(x-3)}=-\infty \) nên $x=2$ là TCĐ 

Vì \(x\in [-2;2)\) nên không tồn tại \(\lim\limits_{x\to +\infty }y\) nên đths không có TCN 

Còn $x=3$ không thể là TCĐ vì tại $x=3$ thì $\sqrt{4-x^2}$ không tồn tại .

1 have got nothing in common

2 put up with his rude

3 everything except the

4 on the verge of speeding 

5 has not changed since

6 the moment we arrive

7 set his heart to become

8 is on the tip of my tongue

9 She was caught to smoke in the bathroom

10 He congratulated them on winning the race

11 had arrived on time

12 looking forward to using

13 prevented the visitors from being

14 could have broken into

15 has a extreme command of 

16 expressed their disapproval of