16 nha bạn

CHÚC BẠN HỌC TỐT!!

(2x+2y)²

=4x²+8xy+4y²

Vậy tổng các hệ số khi khai triền là :

4+8+4=16

k nhé bạn !

16 đấy chị mình giảng cho mình rồi

\(\left(3x^2-2y\right)^3=27x^6-54x^4y+36x^2y^2-8y^3\)

(2x+3y)³=8x³+36x²y+54xy²+27y³

\(\left(2x+3y\right)^3=8x^3+36x^2y+54xy^2+27y^3\)

   (2\(x\) + 3y)²

= (2\(x\) + 3y).(2\(x\) + 3y)

= 4\(x\)² + 6\(xy\) + 6\(xy\) + 9y²

= 4\(x^2\) + 12\(xy\) + 9y²

\(4x^2+12xy+9y^2\)

4�2x2 + 12$xy$ + 9y2

Trả lời:

a, \(\left(\frac{1}{2}x-3\right)^2=\left(\frac{1}{2}x\right)^2-2.\frac{1}{2}x.3+3^2=\frac{1}{4}x^2-3x+9\)

b, \(\left(2x^2+3y\right)^2=\left(2x^2\right)^2+2.2x^2.3y+\left(3y\right)^2=4x^4+12x^2y+9y^2\)

c, \(\left(3x-7y\right)^2=\left(3x\right)^2-2.3x.7y+\left(7y\right)^2=9x^2-42xy+49y^2\)

Bài 1:

\(\left(3x+4\right)^2=9x^2+24x+16\)

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

\(\left(x-3\right)^2=x^2-6x+9\)

\(\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25\)

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

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

\(x^2-2=\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)

\(4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)\)

Bài 3:

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

\(x^2-10x+25=\left(x-5\right)^2\)

\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

\(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)

\(\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25\)

help me

\(\left(x-\frac{1}{2}\right)^2=x^2-x+\frac{1}{4}\)

(2x - 3y)² = 4x² - 12xy + 9y²