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 

a) vì trong tam giác cân đường cao đồng thời là đường trung tuyến nên AH là đường trung tuyến nên BH = CH

b) ta có BH=CH =1/2BC = 3(cm)

ΔABH vuông tại H

Áp dụng định lý Pi-ta-go, ta có:

AH²+BH²=AB²

⇒ AH² = AB²-BH²

⇒ AH² = 5²-3²

⇒ AH²= 16

⇒ AH = 4(cm)

Bài 1:

Thời gian ô tô đi hết quãng đường:

\(t=\dfrac{S}{v}=\dfrac{60}{40}=\dfrac{3}{2}\left(h\right)\)

Vận tốc lúc về là:

\(40+10=50\left(\dfrac{km}{h}\right)\)

Thời gian ô tô đi về:

\(t_2=\dfrac{S_2}{v_2}=\dfrac{60}{50}=\dfrac{6}{5}\left(h\right)\)

a, \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,25    0,25          0,25     0,25

\(m_{Fe}=0,25.56=14\left(g\right)\)

\(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

b, \(m_{FeSO_4}=0,25.162=40,5\left(g\right)\)

khó thật , đọc hỉu j đâu 

\(1+cota+cot^2a+cot^3a\)

\(=1+\dfrac{cosa}{sina}+\dfrac{cos^2a}{sin^2a}+\dfrac{cos^3a}{sin^3a}\)

\(=\left(1+\dfrac{cosa}{sina}\right)\left(1+\dfrac{cos^2a}{sin^2a}\right)\)

\(=\dfrac{sina+cosa}{sina}.\dfrac{sin^2a+cos^2a}{sin^2a}\)

\(=\dfrac{cosa+sina}{sin^3a}\)

Bn ơi!Nếu bài này có hình vẽ thì bn đăng luôn hình nha!

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

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