Xét (I) có 

\(\widehat{BAC}=\dfrac{1}{2}\widehat{BIC}\)

nên \(\widehat{BIC}=2\cdot31^0=62^0\)

Xét (K) có 

\(\widehat{DIE}=\dfrac{1}{2}\cdot\widehat{DKE}\)

nên \(\widehat{DKE}=2\cdot62^0=124^0\)

\(\widehat{BAC}=\dfrac{1}{2}sđ\stackrel\frown{BC}\)

\(\widehat{CBx}=\dfrac{1}{2}sđ\stackrel\frown{BC}\)

Do đó: \(\widehat{BAC}=\widehat{CBx}\)

mà \(\widehat{BOC}=2\cdot\widehat{BAC}\)

nên \(\widehat{BOC}=2\cdot\widehat{CBx}\)

Số tiền Minh đã dùng để mua là:

3000 + 4000 = 7000 (đồng)

Số tiền còn lại của Minh là:

10000 - 7000 = 3000 (đồng)

* Bác bán hàng có thể trả cho Minh:

+ Tờ bạc 2000 đồng và 1000 đồng

+ 3 tờ bạc 1000 đồng

...

Chúc bạn học tốt!! ^^

Chọn D

1 The boys like playing video games

2 My mother doesn't like listen to music

3 You should eat more fresh fish, which makes you smarter

4 He had a headache last night, sho he didn't sleep well

5 Doing exercises makes you healthier

6 My dad enjoys planting trees

The boys like playing video games

 My mother doesn't like listen to music

 You should eat more fresh fish, which makes you smarter

He had a headache last night, sho he didn't sleep well

 Doing exercises makes you healthier

My dad enjoys planting trees

The boys like playing video games

 My mother doesn't like listen to music

 You should eat more fresh fish, which makes you smarter

 He had a headache last night, sho he didn't sleep well

 Doing exercises makes you healthier

 My dad enjoys planting trees

1 will have - will invite

2 goes

3 play

4 won't study

5 does

6 will travel

7 won't come

8 will come

bài 2

1 riding - walking

2 swimming

3 watching

4 tell

5 travel

6 studying

7 seeing

8 go - have

1. She likes playing guitar

2. He likes reading book

3.She likes singing 

4.He likes playing soccer

\(c,x\left(x-20\right)-x+20=0\\ \Leftrightarrow x^2-20x-x+20=0\\ \Leftrightarrow x^2-21x+20=0\\ \Leftrightarrow\left(x-20\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=20\\x=1\end{matrix}\right.\)

\(b,x^2-4+\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow x^2-4+x^2-x-6=0\\ \Leftrightarrow2x^2-x-10=0\\ \Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

a: Ta có: \(x\left(x-20\right)-x+20=0\)

\(\Leftrightarrow\left(x-20\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=1\end{matrix}\right.\)

b: Ta có: \(x^2-4+\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\)

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

CuO + H2 -> Cu + H2O

0.01              0.01

FexOy + yH2 -> xFe + yH2O

Fe            +        2HCl ->     FeCl2 +           H2

\(\)0.02                                                          0.02                                              \(\)

Cu + HCl -> (không phản ứng)

nH2 = 0.02mol => mFe = 1.12g

=> mCu = 1.76 - 1.12 = 0.64g => nCu = 0.01mol

=> mCuO = 0.8g => mFexOy = 2.4 - 0.8 = 1.6g

Ta có: 56x + 16y -> 56x

             1.6g       -> 1.12g

=> \(1.6\times56x=1.12\times\left(56x+16y\right)\)

=> \(26.88x=17.92y\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

=> Fe2O3

\(log_{\sqrt{x}}y=\dfrac{2y}{5}\Rightarrow2log_xy=\dfrac{2y}{5}\) \(\Rightarrow log_xy=\dfrac{y}{5}\)

\(log_{\sqrt[3]{5}}x=\dfrac{15}{y}\Rightarrow3log_5x=\dfrac{15}{y}\Rightarrow log_5x=\dfrac{5}{y}\)

\(\Rightarrow log_xy=\dfrac{1}{log_5x}=log_x5\Rightarrow y=5\)

\(\Rightarrow log_5x=\dfrac{5}{5}=1\Rightarrow x=5\)

\(\Rightarrow x^2+y^2=25+25=50\)

Chọn B