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x^4-5x^2+4=x^4-x^2-(4x^2-4) = x^2(x^2-1)-4(x^2-1)
=(x^2-4)(x^2-1)
=(x-2)(x+2)(x-1)(x+1)
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
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~
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)
\(=x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128\)
\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(=\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\)
\(=\left(x^2+10x\right)^2+2.\left(x^2+10x\right).12+12^2-16\)
\(=\left(x^2+10x+12\right)^2-4^2\)
\(=\left(x^2+10x+12-4\right) \left(x^2+10x +12+4\right)\)
\(=\left(x^2+10x-8\right)\left(x^2+10x+16\right)\)
\(=\left(x^2+10x-8\right)\left(x^2+2x+8x+16\right)\)
\(=\left(x^2+10x-8\right)\left[x\left(x+2\right)+8\left(x+2\right)\right]\)
\(=\left(x^2+10x-8\right)\left(x+2\right)\left(x+8\right)\)
A= x(x+4)(x+6)(x+10) +128
=[(x(x+10)] [(x+4)(x+6)] +128
=(x^2+10)(x^2+10+24)+128
Đặt: x^2+10+12=y
Ta có: A=(y+12)(y-12)+128
=(y^2-12^2)+128
=y^2-12^2+128
=y^2-16
=y^2-4^2
=(y-4)(y+4)
Thay vào bt A ta có:A= ( x^2+10x+12-4)(x^2+10x+12+4)
=(x^2+10x+8)(x^2+10x+16)
=(x^2+10x+8)(x+8)(x+2)x
x . ( x + 4 ) . ( x + 6 ) . ( x + 10 ) + 128
= ( x2 + 10x ) . ( x2 + 10x + 24 ) + 128
đặt x2 + 10x + 12 = y, đa thức đã cho có dạng :
( y - 12 ) . ( y + 12 ) + 128 = y2 - 16 = ( y - 4 ) . ( y + 4 )
= ( x2 + 10x + 16 ) . ( x2 + 10x + 8 ) = ( x + 2 ) . ( x + 8 ) . ( x2 + 10x + 8 )
a) x4 - 4x3 + 8x + 3 b)x3+ 3x2-2
= x4 - (2x3 + 2x3 ) + (2x+6x) + 3 + 4x2 - 4x2 = x3 + 2x2+ x2 - 2 + 2x - 2x
= x4 - 2x3 - x2 - 2x3+4x2 + 2x - 3x2 +6x + 3 = ( x3 + 2x2 -2x) +( x2+ 2x - 2)
= x2(x2 - 2x - 1) - 2x( x2- 2x -1) - 3 ( x2- 2x - 1) =x(x2+2x-2) + (x2 + 2x - 2)
= ( x2 - 2x -1)(x2- 2x -3 ) = (x2+ 2x - 2 ) (x+1)
c) x2-6x2+ 16
= (- 5)x2 + 16
= - ( 5x2 - 16)
= [ 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)