Giúp mình pt thành nhân tử x^6-x^4-x^3-x^2 THANKS
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\(x^6-x^4+2x^3+2x^2\)
\(=x^4.\left(x^2-1\right)+2x^2.\left(x+1\right)\)
\(=x^4.\left(x-1\right).\left(x+1\right)+2x^2.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^5-x^4+2x^2\right)\)
pt đa thức thành nhân tử x*(x+4)*(x+6)*(x+10)+128
bạn nào làm ra cách giải sớm cho mình mình tick cho
= [ x ( x + 10 ) ] [ ( x+4 ) ( x+ 6) +128
=( x2 + 10x ) ( x2 +10x + 24 ) +128
dat : x2 + 10x =a , ta co:
a ( a + 24 ) +128
=a2 + 24a +128
= (a + 12 )2 - 16
= ( a+ 12 -4 ) ( a + 12 + 4)
= ( a +8 ) ( a + 16 )
= ( x2 + 10x +8 )( x2 + 10x + 4)
\(=\left(x^2+2xz+z^2\right)-\left(4y^2-4y+1\right)\)
\(=\left(x+z\right)^2-\left(2y-1\right)^2\)
\(=\left(x+z-2y+1\right)\left(x+z+2y-1\right)\)
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
Đặt \(t=x^2-11x+30\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=t.\left(t-2x\right)-24x^2\)
\(=t^2-2xt-24x^2\)
\(=\left(t^2-2xt+x^2\right)-25x^2\)
\(=\left(t-x\right)-\left(5x\right)^2\)
\(=\left(t-6x\right)\left(t+4x\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
Tham khảo nhé~
\(x^2-3xy-40y^2\)
\(=x^2+5xy-8xy-40y^2\)
\(=x\left(x+5y\right)-8y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-8y\right)\)
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
a) \(x^2-2xy+y^2-xz+yz\)
= \(\left(x-y\right)^2-z\left(x-y\right)\)
= \(\left(x-y\right)\left(x-y-z\right)\)
b) \(x^3+9x^2-4x-36\)
= \(x^3-2x^2+11x^2-22x+18x-36\)
= \(x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\)
= \(\left(x-2\right)\left(x^2+11x+18\right)\)
= \(\left(x-2\right)\left(x^2+2x+9x+18\right)\)
= \(\left(x-2\right)\left(x+2\right)\left(x+9\right)\)
Chuc ban hoc tot
\(\Rightarrow3\left(x^3-8\right)-3x^3-3x=-30\\ \Rightarrow3x^3-24-3x^3-3x=-30\\ \Rightarrow-3x=-6\Rightarrow x=2\)
\(3\left(x-2\right)\left(x^2+2x+4\right)-3x\left(x^2+1\right)=-30\)
\(\Leftrightarrow3x^2-24-3x^3-3x=-30\)
\(\Leftrightarrow x=2\)
x2(x+1)(x3-X2-1)