2 should not go out for a walk

3 will quit it at once

4 must buy a smaller one

5 may be very worried

6 may find another one

7 can understand the lecture well

8 has to leave the building in order

9 ought to come to visit me more

10 must be very happy

19

Từ pt đầu ta có:

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

\(\Leftrightarrow x\left(x-y\right)-2y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

TH1: \(x=y\) thế xuống pt dưới:

\(y^2-y-y^2=1\Rightarrow y=-1\Rightarrow x=-1\)

TH2: \(x=2y\) thế xuống pt dưới:

\(\left(2y\right)^2-2y-y^2=1\Leftrightarrow3y^2-2y-1=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-\dfrac{1}{3}\Rightarrow x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(-1;-1\right);\left(1;2\right);\left(-\dfrac{1}{3};-\dfrac{2}{3}\right)\)

21.

Từ pt đầu:

\(xy+2=2x+y\Leftrightarrow xy-y+2-2x=0\)

\(\Leftrightarrow y\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TH1: \(x=1\) thế xuống pt dưới:

\(2y+y^2+3y=6\Leftrightarrow y^2+5y-6=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-6\end{matrix}\right.\)

TH2: \(y=2\) thế xuông pt dưới

\(4x+4+6=6\Rightarrow x=-1\)

Vậy nghiệm của pt là: \(\left(x;y\right)=\left(1;1\right);\left(1;-6\right);\left(-1;2\right)\)

22.

\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-6xy+y^2+6xy=10\\\left(3x-y\right)\left(10-6xy\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-y\right)^2=10-6xy\\\left(3x-y\right)\left(10-6xy\right)=8\end{matrix}\right.\)

Thế \(10-6xy\) từ pt trên xuống dưới ta được 

\(\left(3x-y\right)\left(3x-y\right)^2=8\)

\(\Leftrightarrow\left(3x-y\right)^3=2^3\)

\(\Leftrightarrow3x-y=2\)

\(\Leftrightarrow y=3x-2\)

Thế vào pt đầu:

\(9x^2+\left(3x-2\right)^2=10\)

\(\Leftrightarrow18x^2-12x-6=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-\dfrac{1}{3}\Rightarrow y=-3\end{matrix}\right.\)

Ai giúp em với ạ

chữ đẹp quá :))

Bài 6:

Xét ΔOAC vuông tại A và ΔOBD vuông tại B có

OA=OB

\(\widehat{AOC}=\widehat{BOD}\)(hai góc đối đỉnh)

Do đó: ΔOAC=ΔOBD

=>OC=OD

Bài 7:

a: Ta có: \(\widehat{DAB}+\widehat{BAC}+\widehat{CAE}=180^0\)

=>\(\widehat{DAB}+\widehat{CAE}+90^0=180^0\)

=>\(\widehat{DAB}+\widehat{CAE}=90^0\)

mà \(\widehat{DAB}+\widehat{DBA}=90^0\)

nên \(\widehat{DBA}=\widehat{CAE}\)

Xét ΔABD vuông tại A và D và ΔCAE vuông tại E có

AB=AC

\(\widehat{DBA}=\widehat{EAC}\)

Do đó: ΔABD=ΔCAE

b: ta có: ΔABD=ΔCAE

=>DB=AE và AD=CE

DB+CE=DA+AE=DE

tách nhỏ câu hỏi ra

1, When does he go to school

2, What does it mix

3, Who enjoys playing the piano

      What does Lan enjoy (doing)

4, Who stipped on her cappuccino

   What did Katie stip on

5, Why were you snippy with him

6, Who are the symbol of........

  Why are evergreen trees a symbol of fertility

7, How long will the show last

8,What have they spent on their new laptop

9, What is Tyara buying

10, Where were the kids playing yesterday

11, What have the scientists done before the vaccine was released 

12, How wide is the pool

13, How far is it from your house to school

14, What must a doctor put first

15, Who has practiced violin...................

       What has Vivian practiced hard in..........

      Why has Vivian practiced Violin hard?

1, Who will captain the team...............

2, What is your sis's job

3, What do you often listen to

a: Ta có: \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Leftrightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Leftrightarrow-9x=-18\)

hay x=2

b: ta có: \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+x-6x^2+5x-3=0\)

\(\Leftrightarrow17x=-5\)

hay \(x=-\dfrac{5}{17}\)

c: Ta có: \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2-12-14x+13=0\)

\(\Leftrightarrow8x^3-8x=0\)

\(\Leftrightarrow8x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

a) \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Rightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Rightarrow-9x=-18\)

\(\Rightarrow x=2\)

b) \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Rightarrow x^3+6x^2+12x+8-x^3+x=6x^2-5x+3\)

\(\Rightarrow18x=-5\)

\(\Rightarrow x=-\dfrac{5}{18}\)

c) \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Rightarrow8x^3-12x^2+6x-1+12x^2-12=14x-13\)

\(\Rightarrow8x^3-8x=0\)

\(\Rightarrow8x\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)