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11. 

\(tan\left(x-\pi\right)=-tan\left(\pi-x\right)=tanx\)

12.

\(sinx+sin3x=2sin\dfrac{x+3x}{2}.cos\dfrac{x-3x}{2}=2sin2x.cos\left(-x\right)=2sin2x.cosx\)

15.

\(x^2+y^2-2x+4y-4=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=9\)

\(\Rightarrow\) Tâm \(I=\left(1;-2\right)\), bán kính \(R=3\)

a: Ta có: \(\sqrt{x+2}=3x-4\)

\(\Leftrightarrow9x^2-24x+16-x-2=0\)

\(\Leftrightarrow9x^2-25x+14=0\)

\(\text{Δ}=\left(-25\right)^2-4\cdot9\cdot14=121\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{25-11}{18}=\dfrac{7}{18}\left(loại\right)\\x_2=\dfrac{25+11}{18}=2\left(nhận\right)\end{matrix}\right.\)

x : 3 dư 2

x : 5 dư 1

→ x + 4 chia hết cho 3 và 5

→ x + 4 € BC ( 3, 5 )

Ta có: 3 . 5 = 15

→ BC ( 3, 5 ) = B ( 15 ) = {0;15;30;45;...}

Dựa vào các điều kiện trên, ta kết luận: Vậy x € { 15;30 }

Áp dụng công thức : a = bq + r , ta có

x : 4 = 16 dư 3

x = 16 . 4 + 3

x = 67

x : 5 = 22 dư 1

x = 22 . 5 + 1

x = 111

x : 4 = 16 (dư 3)

      x =16 x 4 + 3

      x = 67.

x :5 = 22 (dư 1 )

    x = 22 x 5 +1

     x= 111

Kẻ AH⊥BC

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

\(m_{H_2O}=1,62\left(g\right)\Rightarrow n_{H_2O}=0,09\left(mol\right)\Rightarrow n_H=0,18\left(mol\right);m_H=0,18.1=0,18\left(g\right)\\ n_{CO_2}=\dfrac{2,64}{44}=0,06\left(mol\right)\Rightarrow n_C=n_{CO_2}=0,06\left(mol\right);m_C=0,06.12=0,72\left(g\right)\\ Vây:m_C+m_H=0,72+0,18=0,9< 1,38\\ \Rightarrow X.có.chứa.O\\ m_O=1,38-0,9=0,48\left(g\right);n_O=\dfrac{0,48}{16}=0,03\left(mol\right)\\ Đặt.X:C_aH_bO_c\left(a,b,c:nguyên,dương\right)\\ Ta.có:a:b:c=0,06:0,18:0,03=2:6:1\\ \Rightarrow CTĐG:C_2H_6O\\ M_X=23.2=46\left(\dfrac{g}{mol}\right)=M_{C_2H_6O}\\ \Rightarrow X:C_2H_6O\)

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

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

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 

Bài 2:

a: Ta có: \(M=2x\left(2x^3-3x\right)-x^2\left(3x^2-2\right)-x^2\left(x^2-4\right)\)

\(=4x^4-6x^2-3x^4+2x^2-x^4+4x^2\)

=0

b: Ta có: \(N=x\left(y^2-x\right)-y\left(xy-x^2\right)-x\left(xy-x-1\right)\)

\(=xy^2-x^2-xy^2+x^2y-x^2y+x^2+x\)

\(=x\)