Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
Bài làm :
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
9x2-2015x+2006
= 9x2-9x-2006x+2006
= (9x2-9x)-(2006x-2006)
= 9x(x-1)-2006(x-1)
= (x-1) (9x-2006)
Chúc học tốt nhé!
a) Hình như phân tích không được
b) \(2x^4+5x^3+13x^2+25x+15\)
\(=x^3+1+2x^4+2x^3+13x^2+13x+12x+12+2+2x^3\)
\(=\left(x^3+1\right)+\left(2x^4+2x^3\right)+\left(13x^2+13x\right)+\left(12x+12\right)+2\left(1+x^3\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x^3\left(x+1\right)+13x\left(x+1\right)+12\left(x+1\right)+2\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x^3+13x+12+2x^2-2x+2\right)\)
\(=\left(x+1\right)\left(3x^2+10x+15+2x^3\right)\)
\(=\left(x+1\right)\left[x^2\left(2x+3\right)+5\left(2x+3\right)\right]\)
\(=\left(x+1\right)\left(x^2+5\right)\left(2x+3\right)\)
Ta có : 5x(x - 2y) + 2(2y - x)2
= 5x(x - 2y) + 2(x - 2y)2 (vì (2y - x)2 = (x - 2y)2 )
= (x - 2y)[5x + 2(x - 2y)]
= (x - 2y)(5x + 2x - 4y)
= (x - 2y)(7x - 4y)
b) 7x(y - 4)2 - (4 - y)3
= 7x(y - 4)2 - (4 - y)2(4 - y)
= 7x(y - 4)2 - (y - 4)2(4 - y)
= (y - 4)2(7x - 4 + y)
c) (4x - 8)(x2 + 6) - (4x - 8)(x + 7) + 9(8 - 4x)
= (4x - 8)(x2 + 6) - (4x - 8)(x + 7) - 9(4x - 8)
= (4x - 8)(x2 + 6 - x - 7 - 9)
= 2(x - 4)(x2 - x - 10)
ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)
4x4-37x2+9
=4x4-12x2+9-25
=(2x2-3)2-25
=(2x2-3-5)(2x2-3+5)
=(2x2-8)(2x2+2)
=2.(x2-4).2.(x2+1)
=4(x-2)(x+2)(x2+1)
x^8+x^4+1
=x8+2x4+1-x4
=(x4+1)2-x4
=(x4-x2+1)(x4+x2+1)
=(x4-x2+1)(x4+2x2+1-x2)
=(x4-x2+1)[(x2+1)2-x2]
=(x4-x2+1)(x2-x+1)(x2+x+1)