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1) =\(x^7-x+x^2+x\)+1
=\(x\left(x^6-1\right)+\left(x^2+x+1\right)\)
=\(x\left(x^3-1\right)\left(x^3+1\right)\)\(+\left(x^2+x+1\right)\)
=x(x^3+1)(x-1)(x^2+x+1)+(x^2+x+1)
=[(x^4+x)(x-1)+1](x^2+x+1)
=(x^5-x^4+x^2-x)(x^2+x+1)
Trả lời:
1, x7 + x2 + 1
= x7 + x2 + 1 + x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - x
= ( x7 + x6 + x5 ) - ( x6 + x5 + x4 ) + ( x4 + x3 + x2 ) - ( x3 + x2 + x ) + ( x2 + x + 1 )
= x5 ( x2 + x + 1 ) - x4 ( x2 + x + 1 ) + x2 ( x2 + x + 1 ) - x ( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x5 - x4 + x2 - x + 1 )
b, x8 + x7 + 1
= x8 + x7 + 1 + x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - x
= ( x8 + x7 + x6 ) - ( x6 + x5 + x4 ) + ( x5 + x4 + x3 ) - ( x3 + x2 + x ) + ( x2 + x + 1 )
= x6 ( x2 + x + 1 ) - x4 ( x2 + x + 1 ) + x3 ( x2 + x + 1 ) - x ( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
Sửa đề: \(x^2+2xy+y^2+2x+2y-15\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)
Đặt \(x+y=t\)
\(\Rightarrow t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2\)
\(=\left(t+1-4\right)\left(t+1+4\right)\)
\(=\left(t-3\right)\left(t+5\right)\)
\(=\left(x+y-3\right)\left(x+y+5\right)\)
\(1,\)
\(\left(x^2-x+2\right)^4-3x^2\left(x^2-x+2\right)^2+2x^4\)
Đặt: \(\left(x^2-x+2\right)^2=n\)
\(\left(x^2-x+2\right)^4-3x^2\left(x^2-x+2\right)^2+2x^4\)
\(=n^2-3x^2n+2x^4\)
\(=n\left(n-2x^2\right)-x^2\left(n-2x^2\right)\)
\(=\left(n-2x^2\right)\left(n-x^2\right)\)
Thay \(\left(x^2-x+2\right)^2\)ta có:
\(=[\left(x^2-x+2\right)^2-2x^2][\left(x^2-x+2\right)^2-x^2]\)
\(=[\left(x^2-x+2\right)^2-2x^2]\left(x^2-x+2-x\right)\left(x^2-x+2+x\right)\)
\(=[\left(x^2-x+2\right)^2-2x^2]\left(x^2-2x+2\right)\left(x^2+2\right)\)
\(2,\)
\(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4\)
\(=3\left(-x^2+2x+3\right)^4+x^2\left(-x^2+2x+3\right)^2-27x^2\left(-x^2+2x+3\right)^2-9x^4\)
\(=\left(-x^2+2x+3\right)^2[3\left(-x^2+2x+3\right)^2+x^2]-9x^2[\left(-x^2+2x+3\right)^2+x^2]\)
\(=[3\left(-x^2+2x+3\right)^2+x^2][\left(-x^2+2x+3\right)^2-9x^2]\)
\(x^2+2xy+y^2+2x+2y-15\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)
\(=\left(x+y+1\right)^2-4^2=\left(x+y-3\right)\left(x+y+5\right)\)
\(x^2+2xy+y^2+2x+2y-15\)
\(=x^2+2xy+y^2+2x+2y+1-16\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)
Đặt \(x+y=t\)
\(\Rightarrow t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2\)
\(=\left(t+1-4\right)\left(t+1+4\right)\)
\(=\left(t-3\right)\left(t+5\right)\)
\(=\left(x+y-3\right)\left(x+y+5\right)\)
phân tích đa thức thành nhân tử 4x^3+6x^2+4x+1
= 4x^3+6x^2+4x+1
= (2x+1)(2x^2+2x+1)
nha bạn chúc bạn học tốt nha
\(4x^3+6x^2+4x+1\)
\(=4x^3+4x^2+2x^2+2x+2x+1\)
\(=\left(4x^3+4x^2+2x\right)+\left(2x^2+2x+1\right)\)
\(=2x\left(2x^2+2x+1\right)+\left(2x^2+2x+1\right)\)
\(=\left(2x^2+2x+1\right)\left(2x+1\right)\)
phân tích đa thức thành nhân tử 6x^3+x^2+x+1
= 6x^3+x^2+x+1
= (2x+1)(3x^2-x+1)
chúc bạn học tốt nha
\(x^3+3x^2-10x-24\)
\(=x^3-3x^2+6x^2-18x+8x-24\)
\(=x^2\left(x-3\right)+6x\left(x-3\right)-8\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+6x-8\right)\)
\(=\left(x-3\right)\left(x^2+6x+9-1\right)\)
\(=\left(x-3\right)[\left(x-3\right)^2-1]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+4\right)\)
\(2x^3-11x^2+10x+8\)
\(=2x^3-4x^2-7x^2+14x-4x+8\)
\(=2x^2\left(x-2\right)-7x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2-7x-4\right)\)
\(=\left(x-2\right)[2x\left(x-4\right)+\left(x-4\right)]\)
\(=\left(x-2\right)\left(x-4\right)\left(2x+1\right)\)
Ta có: \(^{3x^3-4x^2+13x-4}\) = \(3x^3-x^2-3x^2+x+12x-4\)
= \(3x^2\left(x-\frac{1}{3}\right)-3x\left(x-\frac{1}{3}\right)+12\left(x-\frac{1}{3}\right)\)
= \(\left(3x^2-3x+12\right)\left(x-\frac{1}{3}\right)\)
= \(3\left(x^2-x+4\right)\left(x-\frac{1}{3}\right)\)
c) (xy^2+1)^2
d) (1/3-y^4)^2
e) (1/2a-2b^2)^2
f) (5-x)^2
c) 2xy2 + x2 y4 + 1
= (xy2)2 + 2xy2 .1 + 12
=(xy2 +1)2
d) 1/9 - 2/3 y4 + y8
= (1/3)2 - 2 . 1/3 y4 + (y4)2
=(1/3 - y4)2
e) 1/4 a2 - 2ab2 + 4b4
= (1/2 a)2 - 2 1/2 a . 2b + (2b2)2
=(1/2 a - 2b2)2
f) 25 - 10x + x2
= x2 + 2 . 5x + 52
= (x+5)2