Bài 7:

a. \(\left(x-3\right)^2-\left(5-2x\right)^2=\left(x-3-5+2x\right)\left(x-3+5-2x\right)=\left(3x-8\right)\left(2-x\right)\)

b. \(\left(x+y\right)^2-x^2+4xy-4y^2=\left(x+y\right)^2-\left(x-2y\right)^2=\left(x+y-x+2y\right)\left(x-y+x-2y\right)=3y\left(2x-3y\right)\)

c. \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]=2y\left(3x^2+y^2\right)\)

d. \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

1.Where are the chairs?

2.Can you use chopstick?

3.Where does Nam live?

4.What subject do you like?

5.What is her telephone number?

\(\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)

\(=\left(1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1+\sqrt{x}-\sqrt{x}-x=1-x\)

\(\sqrt{19-8\sqrt{3}}=\sqrt{16-2.4\sqrt{3}+3}=\sqrt{\left(4-\sqrt{3}\right)^2}=4-\sqrt{3}\)

15. has just taken

16. Have you done/finished

1.A

2.C

3.D

tiền là trước chứ

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

Vì \(\left|4x-3\right|\ge0\) nên \(225-\left|4x-3\right|\le225\)

Dấu "=" xảy ra khi và chỉ khi \(4x-3=0\Leftrightarrow x=\dfrac{3}{4}\)

Vậy \(A_{max}=225\) khi \(x=\dfrac{3}{4}\)

Regardless of her her saying that she would come, I don't think she ever will.