Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(5x+2y\right)\left(3x-8y\right)=\left[\left(4x-3y\right)+\left(x+5y\right)\right]\left[\left(4x-3y\right)-\left(x+5y\right)\right]\)
\(=\left(4x-3y\right)^2-\left(x+5y\right)^2\)
(x-1)^2-2(x-1)(2y-1)+(2y-1) = [ x-1 - (2y-1)]2 = ( x-1-2y+1)2 = ( x-2y)2
a) 25x² - 10xy + y²
= (5x)² - 2.5x.y + y²
= (5x - y)²
b) 4/9 x² + 20/3 xy + + 25y²
= (2/3 x)² + 2.2/3 x.5y + (5y)²
= (2/3 x + 5y)²
c) 9x² - 12x + 4
= (3x)² - 2.3x.2 + 2²
= (3x - 2)²
d) Sửa đề: 16u²v⁴ - 8uv² + 1
= (4uv²)² - 2.4uv².1 + 1²
= (4uv² - 1)²
a) \(\frac{1}{9}x^4-2x^2y+9y^2=\left(\frac{1}{3}\right)^2\left(x^2\right)^2-2x^2y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2\right)^2-2\frac{1}{3}x^23y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2-3y\right)^2\)
b) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
\(\frac{1}{9}x^4-2x^2y+9y^2\)
\(=\left(\frac{1}{3}x^2\right)^2-2\times\frac{1}{3}x^2\times3y+\left(3y\right)^2\)
\(=\left(\frac{1}{3}x^2-3y\right)^2\)
\(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2\times5x\times2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
a) x^3-3x^2+3x-1
=x3-3x2.1+3x.12-13
=(x-1)3
b)16+8x+x^2
=42+2.4.x+x2
=(4+x)2
c) 3x^2+3x+1+x^3
=x3+3x2.1+3x.12+13
=(x+1)3
d)1-2y+y^2
=1-2.1.y+y2
=(1-y)2
\(\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.2\left(3x-2y\right)+2^2\\ =\left(3x-2y+2\right)^2\)
Áp dụng HĐT số 1 : \(A^2+2AB+B^2=\left(A+B\right)^2\)