1: \(\left(2x+1\right)^2=\left(2x\right)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\)

2: \(\left(3x+\frac13y\right)^2=\left(3x\right)^2+2\cdot3x\cdot\frac13y+\left(\frac13y\right)^2=9x^2+2xy+\frac19y^2\)

3: \(\left(xy-\frac12y^2\right)^2\)

\(=\left(xy\right)^2-2\cdot xy\cdot\frac12y^2+\left(\frac12y^2\right)^2\)

\(=x^2y^2-xy^3+\frac14y^4\)

4: \(\left(-\frac32x+2\right)^2=\left(\frac32x-2\right)^2\)

\(=\left(\frac32x\right)^2-2\cdot\frac32x\cdot2+2^2=\frac94x^2-6x+4\)

5: \(\left(5x-\frac14y\right)\left(5x+\frac14y\right)=\left(5x\right)^2-\left(\frac14y\right)^2=25x^2-\frac{1}{16}y^2\)

6: \(\left(x^2+\frac23x\right)\left(\frac23x-x^2\right)=\left(\frac23x\right)^2-\left(x^2\right)^2=\frac49x^2-x^4\)

7: \(\left(2x+y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

\(=8x^3+12x^2y+6xy^2+y^3\)

8: \(\left(1-\frac13y\right)^3=1^3-3\cdot1^2\cdot\frac13y+3\cdot1\cdot\left(\frac13y\right)^2-\left(\frac13y\right)^3\)

\(=1-y+\frac13y^2-\frac{1}{27}y^3\)

9: \(\left(x+4\right)\left(x^2-4x+16\right)\)

\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)

\(=x^3+4^3=x^3+64\)

10: \(\left(\frac45-5x\right)\left(\frac{16}{25}+4x+25x^2\right)=\left(\frac45-5x\right)\left\lbrack\left(\frac45\right)^2+\frac45\cdot5x+\left(5x\right)^2\right\rbrack\)

\(=\left(\frac45\right)^3-\left(5x\right)^3=\frac{64}{125}-125x^3\)

11: \(\left(1-x-y\right)^2=\left(x+y-1\right)^2=x^2+y^2+1+2xy-2x-2y\)

12: \(\left(x+y+z\right)\left(x-y-z\right)=x^2-\left(y+z\right)^2\)

\(=x^2-y^2-z^2-2yz\)

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Dỗi đó hả

Part 1:

1. with

2. does

3. after

4. if

5. where

6. I won't

Part 2:

1B 2F 3A 4E 5C 6D

Part 3:

1. invited

2. will play

3. have just won

4. be cleaned

5. to take

6. beautifully

7. impression

Part 4:

1. read -> reading

2. disappointing -> disappointed

3. environment -> environmental

4. who -> that

Part 5:

1. the bad weather

2. are made to study

3. were good at learning

4. going to the English

n = 4 nha bạn

Chúc bạn học tốt

\(\frac{a+1}{2a}=15\left(ĐKXĐ:x\ne0\right)\)

\(\Rightarrow a+1=30a\)

\(\Leftrightarrow x=\frac{1}{29}\left(TMĐK\right)\)

\(\Rightarrow S=\left\{\frac{1}{29}\right\}\)

10B

x=A*cos(\(\omega t+\varphi\))

x=5*cos(\(\omega t\))

=>A=5

11A:

Biên độ dao động bằng quỹ đạo chia đôi

=>A=MN/2=15cm

12C

\(y=-10\cdot cos\left(4\Pi t-\dfrac{pi}{4}\right)=10\cdot cos\left(4\Pi t+\dfrac{pi}{4}\right)\)

=>Pha dao động ban đầu là pi/4

13C

\(y=-8cos\left(2t+\dfrac{pi}{2}\right)=8\cdot\left[-cos\left(2t+\dfrac{pi}{2}\right)\right]\)

\(=8\cdot cos\left(2t-\dfrac{pi}{2}\right)\)

=>Pha dao động ban đầu là -pi/2

mình cũng thấy khó zl ấy ạ =))