Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
j: \(\dfrac{10x^3-19x^2-4x+4}{2x+1}\)
\(=\dfrac{10x^3+5x^2-24x^2-12x+8x+4}{2x+1}\)
\(=5x^2-12x+4\)
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a) \(3x^2-6x\)
\(=3x\left(x-2\right)\)
b) \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-y-1\right)\left(x+y-1\right)\)
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
2 x 2 + 4 x + 8 x 3 - 3 x 2 - x + 3 : P = x 3 - 8 x + 1 x - 3
\(a,x^2-4xa+4a^2-81y^2=\left(x-2a\right)^2-\left(9y\right)^2=\left(x-2a-9y\right)\left(x-2a+9y\right)\\ b,3x^2-8x+4=\left(3x^2-6x\right)-\left(2x-4\right)=3x\left(x-2\right)-2\left(x-2\right)=\left(x-2\right)\left(3x-2\right)\)
Bài 1:
a)2x2+4x-70
=2(x2+2x-35)
=2(x2+7x-5x-35)
=2[x(x+7)-5(x+7)]
=2(x-5)(x+7)
b)x3-5x2+8x-4
=x3-4x2+4x-x2+4x-4
=x(x2-4x+4)-(x2-4x+4)
=(x2-4x+4)(x-1)
=(x-2)2(x-1)
c)x2-10x+16
=x2-2x-8x+16
=x(x-2)-8(x-2)
=(x-8)(x-2)
Bài 2:
\(\frac{8x-8x^3-10x^2+3x^4-5}{3x^2-2x+1}=\frac{\left(x^2-2x-5\right)\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)
bài 2 ghi rõ tí ko hỉu mấy