Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)
(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)
(*)\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+1+x\right)\left(x^2+1-x\right)\)
(*)\(x^4-4x^2+3=x^4-x^2-3x^2+3=x^2\left(x^2-1\right)-3\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)
a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)
b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
c, \(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
d, \(x^4-4x^2+3=x^4-4x^2+4-1=\left(x^2-2\right)^2-1=\left(x^2-1\right)\left(x^2-3\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
Để tính các biểu thức trên, ta sẽ áp dụng quy tắc nhân đa thức.
a) 2xy(3x+1) = 6x^2y + 2xy
b) -6x^2y(4x-5) = -24x^3y + 30x^2y
c) -3x^2(4x^2y-6xy) = -12x^4y + 18x^3y
d) 1/2xy^2(2x+3) = xy^2 + 3/2xy^2
e) 8x^2y^2(1/4xy-1/2x^2) = 2xy - 4x^2y^2
f) 5x(x^2+3x+1) = 5x^3 + 15x^2 + 5x
g) -1/2x^2y(2xy+6) = -x^3y - 3x^2y
a) Ta có: \(3x^2\cdot\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
b) Ta có: \(\left(4xy+3y-5\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^2y\)
c) Ta có: \(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)
\(=12x^2+15x-8x-10-12x^2+6x\)
\(=13x-10\)
d) Ta có: \(\left(3x-5\right)\left(x^2-5x+7\right)\)
\(=3x^3-15x^2+21x-5x^2+25x-35\)
\(=3x^3-20x^2+46x-35\)