4x^4-21x^2y^3+y^4
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=4x^4+21x^2y^2+y^4-25x^2y^2\)
\(\left(2x^2+y^2\right)-\left(5xy\right)^2\)
\(\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)
4x4 - 21 x2y2 + y4
= (4x4 + 4x2y2 + y4) - 25x2y2
= [(2x2)2 + 2x2 . 2 . y2 + (y2)2] - 25x2y2
= (2x2 + y2) - 25x2y2
= (2x2 + y2 - 5xy) (2x2 + y2 + 5xy)
a) x2 - 7xy - 18y2
= x2 + 2xy - 9xy - 18y2
= x(x + 2y) - 9y(x + 2y)
= (x - 9y)(x + 2y)
b) 4x2 + 8x - 5
= 4x2 - 2x + 10x - 5
= 2x(2x - 1) + 5(2x - 1)
= (2x + 5)(2x - 1)
c) 4x4 - 21x2y2 + y4
= (4x4 + 4x2y2 + y4) -25x2y2
= (2x2 + y2) - (5xy)2
= (2x2 + 5xy + y2)(2x2 - 5xy + y2)
= \(2\left(x^2+\frac{5}{2}xy+\frac{y^2}{2}\right)2\left(x^2-\frac{5}{2}xy+\frac{y^2}{2}\right)\)
= \(4\left[\left(x+\frac{5}{4}y\right)^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\left[\left(x-\frac{5}{4}\right)y^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\)
\(=4\left(x+\frac{5}{4}y-\frac{\sqrt{17}}{4}y\right)\left(x+\frac{5}{4}y+\frac{\sqrt{17}}{4}y\right)\left(x-\frac{5}{4}y-\frac{\sqrt{17}y}{4}\right)\left(x-\frac{5}{4}y+\frac{\sqrt{17y}}{4}\right)\)
4x4 - 21 x2y2 + y4
= (4x4 + 4x2y2 + y4) - 25x2y2
= [(2x2)2 + 2x2 . 2 . y2 + (y2)2] - 25x2y2
= (2x2 + y2) - 25x2y2
= (2x2 + y2 - 5xy) (2x2 + y2 + 5xy)
Ấn nhầm :v
a) \(4x^4-21x^2y^2+y^4\)
\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)
\(=\left(2x^2-y^2\right)^2-\left(5xy\right)^2\)
\(=\left(2x^2-5xy-y^2\right)\left(2x^2+5xy-y^2\right)\)
b) \(x^5-5x^3+4x\)
\(=x^5-4x^3-x^3+4x\)
\(=x^3\left(x^2-4\right)-x\left(x^2-4\right)\)
\(=\left(x^2-4\right)\left(x^3-x\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x^2-1\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
4x4 - 21 x2y2 + y4
= (4x4 + 4x2y2 + y4) - 25x2y2
= [(2x2)2 + 2x2 . 2 . y2 + (y2)2] - 25x2y2
= (2x2 + y2) - 25x2y2
= (2x2 + y2 - 5xy) (2x2 + y2 + 5xy)
b: \(=4x^4+4x^2y^2+y^4-25x^2y^2\)
\(=\left(2x^2+y^2\right)^2-25x^2y^2\)
\(=\left(2x^2-5xy+y^2\right)\left(2x^2+5xy+y^2\right)\)
c: \(=2\cdot x^2\cdot\left(x+1\right)^2-\dfrac{1}{2}x^2\)
\(=x^2\left(2x^2+4x+2-\dfrac{1}{2}\right)\)
\(=x^2\left(2x^2+4x+\dfrac{3}{2}\right)\)
\(=x^2\left(2x^2+x+3x+\dfrac{3}{2}\right)\)
\(=x^2\left[x\left(2x+1\right)+\dfrac{3}{2}\left(2x+1\right)\right]\)
\(=x^2\left(2x+1\right)\left(x+\dfrac{3}{2}\right)\)
Qpapaoaoaoaoaoaoaoaoaoaoaoaoao