\(\dfrac{19}{20}-x=\dfrac{8}{5}+\dfrac{3}{4}\)
\(\dfrac{19}{20}-x=\dfrac{32}{20}+\dfrac{15}{20}\)
\(\dfrac{19}{20}-x=\dfrac{47}{20}\)
\(x=\dfrac{19}{20}-\dfrac{47}{20}\)
\(x=\dfrac{-28}{20}=\dfrac{-7}{5}\)

\(\dfrac{19}{20}-x=\dfrac{8}{5}+\dfrac{3}{4}\\ \dfrac{19}{20}-x=\dfrac{47}{20}\\ x=\dfrac{19}{20}-\dfrac{47}{20}\\ x=-\dfrac{28}{20}\\ x=-\dfrac{7}{5}.\)

5.

\(sin\left(60^o+2x\right)=-1\)

\(\Leftrightarrow60^o+2x=-90^o+k.360^o\)

\(\Leftrightarrow x=-75^o+k.180^o\)

6.

\(sin\left(2x+1\right)=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=arcsin\dfrac{1}{3}+k2\pi\\2x+1=\pi-arcsin\dfrac{1}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arcsin\dfrac{1}{3}-\dfrac{1}{2}+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}arcsin\dfrac{1}{3}-\dfrac{1}{2}+k\pi\end{matrix}\right.\)

Cách làm : 

sina = \(\dfrac{1}{2}\) ⇔ \(\left[{}\begin{matrix}a=\dfrac{\pi}{6}+k2\pi\\a=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

sina = \(-\dfrac{\sqrt{3}}{2}\) ⇔ \(\left[{}\begin{matrix}a=-\dfrac{\pi}{3}+k2\pi\\a=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)

sina = 1 ⇔ \(a=\dfrac{\pi}{2}+k.2\pi\)

sina = 0 ⇔ \(a=k\pi\)

sina = -1 ⇔ \(a=-\dfrac{\pi}{2}+k.2\pi\)

sina = \(\dfrac{1}{3}\) ⇔ \(\left[{}\begin{matrix}a=arcsin\left(\dfrac{1}{3}\right)+k2\pi\\a=\pi-arcsin\left(\dfrac{1}{3}\right)+k2\pi\end{matrix}\right.\)

Với a là một đa thức xác định trên R

3.

\(sin\left(\dfrac{x}{2}-45^o\right)=1\)

\(\Leftrightarrow\dfrac{x}{2}-45^o=90^o+k.360^o\)

\(\Leftrightarrow x=270^o+k.720^o\)

4.

\(sin\left(3x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow3x+\dfrac{\pi}{3}=k\pi\)

\(\Leftrightarrow3x=-\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{9}+\dfrac{k\pi}{3}\)

C

B

A

A

D

B

TL:

C

B

A

A

D

B

tự luận tự làm đi nha

H1: x = 360^o - 130^o - 60^o - 82^o = 88^o

H2: x = 360^o - 90^o - 90^o - 72^o = 108^o

H3: x = 360^o - 90^o - 115^o - 70^o = 85^o

H4: 2x = 360^o - 71^o - 105^o = 184^o

=> x = 184^o : 2 = 62^o

Lời giải:
Gọi $I(a,b)$ là tâm đường tròn

$(I)$ tiếp xúc với $(d)$ nên: \(R=d(I,(d))=\frac{|a-b+1|}{\sqrt{2}}(*)\)

Mặt khác: 

\(\overrightarrow{AB}=(6,-2)\)

\(H(9,4)\) là trung điểm $AB$. \(\overrightarrow{HI}=(a-9,b-4)\)

\(\overrightarrow{HI}\perp \overrightarrow{AB}\Rightarrow 6(a-9)-2(b-4)=0\)

\(\Leftrightarrow 3a-b=23\)

Thay vô $(*)$ thì $R=\frac{|24-2a|}{\sqrt{2}}$

Ta cũng có \(R=IA=\sqrt{(a-6)^2+(b-5)^2}=\sqrt{(a-6)^2+(3a-23-5)^2}\)

\(=\sqrt{10a^2-180a+820}\)

Vậy: \(\frac{|24-2a|}{\sqrt{2}}=\sqrt{10a^2-180a+820}\)

$\Leftrightarrow (24-2a)^2=2(10a^2-180a+820)$

$\Leftrightarrow 16a^2-264a+1064=0$

$\Leftrightarrow 2a^2-33a+133=0$

$\Leftrightarrow a=\frac{19}{2}$ hoặc $a=7$

Đến đây bạn tìm được tâm hình tròn, biết bán kính thì sẽ tìm được pt đường tròn.

9 C

10 A

11 A

12 C

13 B

14 D

15 C 

16 D

17 A

18 D

19 B

20 C

21 A

22 A

23 B

24 B

B

1 In spite of being young, she performs excellently

2 The problen of energy shortage will be solved by using solar energy