Phân tích đa thức ra nhân tử: \(y^4-16y^2\)
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\(a,=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,=\left(x+2\right)^2-16y^2=\left(x+4y+2\right)\left(x-4y+2\right)\)
\(16x^2+y^2+4y-16y-8xy\)
\(=\left(16x^2-8xy+y^2\right)+4y-16y\)
\(=\left(4x+y\right)^2-12y\)
\(=\left(4x+y-\sqrt{12y}\right)\left(4x+y-\sqrt{12y}\right)\)
P/S : Sai thì thôi nha!
\(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(x^4-16y^4\)
\(=\left(x^2\right)^2-\left(4y^2\right)^2\)
\(=\left(x^2-4y^2\right)\left(x^2+4y^2\right)\)
\(=\left(x-2y\right)\left(x+2y\right)\left(x^2+4y^2\right)\)
Chúc bạn học tốt.
\(54x^3+16y^3=2\left(27x^3+8y^3\right)\)
\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2\left(3x+2y\right)\left[\left(3x\right)^2-3x.2y+\left(2y\right)^2\right]\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
16y2 - 4x2 - 12x - 9 = 16y2 - (4x2 + 12x + 9) = 16y2 - (2x + 3)2 = (4y - 2x - 3)(4y + 2x + 3)
\(x^3+4x^2+4x-16y^2\)
\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)
\(=x^2.\left(x+2\right)+2x.\left(x+2\right)-16y^2\)
\(=\left(x+2\right).\left(x^2+2x\right)-16y^2\)
\(=x.\left(x+2\right).\left(x+2\right)-\left(4y\right)^2\)
\(=x.\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)
\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)
Tham khảo nhé~
nếu đưa vô căn phải có điều kiện là x > 0
\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left(x\sqrt{x}+2\sqrt{x}\right)^2-\left(4y\right)^2=\left(x\sqrt{x}+2\sqrt{x}-4y\right)\left(x\sqrt{x}+2\sqrt{x}+4y\right)\)
2.\(\left(x^2-16y^2\right)-3x+12y=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)
4.\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
\(16y^2-4x^2-12x-9=16y^2-\left(2x-3\right)^2\)
\(=\left(4y-2x+3\right)\left(4y+2x-3\right)\)
a: \(3x^2-6xy+8x-16y\)
\(=\left(3x^2-6xy\right)+\left(8x-16y\right)\)
\(=3x\left(x-2y\right)+8\left(x-2y\right)\)
\(=\left(x-2y\right)\left(3x+8\right)\)
h: \(9y^2-4x^2+4x-1\)
\(=9y^2-\left(4x^2-4x+1\right)\)
\(=\left(3y\right)^2-\left(2x-1\right)^2\)
\(=\left(3y-2x+1\right)\left(3y+2x-1\right)\)
Ta có: y^4 - 16y^2= (y^2)^2 - (4y)^2
=(y^2 - 4y).(y^2 + 4y)