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bằng phương pháp nào zậy bn????
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a) \(x^3-4x^2-8x+8\)
\(=x^3+8-4x^2-8x\)
\(=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
\(=\left(x+2\right)\left(x^2-6x+9-5\right)\)
\(=\left(x+2\right)\left[\left(x-3\right)^2-5\right]\)
\(=\left(x+2\right)\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)
b) \(1+6x-6x^2-x\)
\(=1-x+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(6x+1\right)\)
a)\(3x^2-8x+4\)
\(=3x^2-2x-6x+4\)
\(=x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
b)\(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
c)\(x^8+98x^4+1\)
\(=\left(x^8+2x^4+1\right)+96x^4\)
\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)
d)\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)\(=\left(x^2+3x-1\right)^2\)
a) x3 + 2x - 3
=x3+x2+3x-x2+x+3
=x(x2+x+3)-1(x2+x+3)
=(x-1)(x2+x+3)
b) x3 - x2 + x + 3
=x3-2x2+3x+x2-2x+3
=x(x2-2x+3)+1(x2-2x+3)
=(x+1)(x2-2x+3)
c) 3x3 - 4x2 + 13x - 4
=3x3-3x2+12-x2-x+4
=3x(x2-x+4)-1(x2-x+4)
=(3x-1)(x2-x+4)
d) 6x3 + x2 + x + 1
=6x3-2x2+2x+3x2-x+1
=2x(3x2-x+1)+1(3x2-x+1)
=(2x+1)(3x2-x+1)
e)bạn phân tích tương tự nhé mk cho đáp án để bạn đổi chiếu nè
=(2x+1)(2x2+2x+1)
a) \(A=x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)
\(=\left[x\left(x+10\right)\right].\left[\left(x+4\right)\left(x+6\right)\right]+128\)
\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
đặt \(x^2+10x+12=t\)khi đó:
\(A=\left(t-12\right)\left(t+12\right)+128\)
\(=t^2-16=\left(t-4\right)\left(t+4\right)\)
bạn thay trở lại nhé
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+6x^3+9x^2-2x^2-6x+1\)
\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x-1\right)^2\)
d) \(x^7+x^2+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
e) \(4x^4+81=4x^4+36x^2+81-36x^2=\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
a) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )