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\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
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^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)
a, x^4+6x^3+11x^2+6x+1
= x^4 + 6x^3 + 9x² + 2x² + 6x + 1
= x^4 + 9x² + 1 + 6x^3 + 2x² + 6x
= x^4 + 9x² + 1² + 2.x².3x + 2.x².1 + 2.3x.1
= (x² + 3x + 1)²
Mình làm được ý a nên tk 1 tk
a, x2-7x-14y+2x
=x(x+2)-7(x-2y)
b, x3-4x2y+4xy2-25x
=x3-4x2y+4xy2-y3-25x+y3
=(x-y)3-25x+y3
a ) = x(x+2) - 7(x+2y)
b) = -4 xy ( x-y) + (x^3-25x) [ câu này mk , chaqcs là làm đúng đâu ]
a) ( x-2 )( x - 4 )( x - 6 )( x -8 ) + 15
= ( x- 2 )( x - 8 )( x - 4)( x- 6 ) + 15
= ( x^2 - 10x + 16 )( x^2 - 10x + 24 ) + 15
Đắt x^2 + x + 16 = y
= y ( y + 8 ) + 15
= y^2 + 8y + 15
= y^2 + 3y + 5y + 15
=y ( y + 3 ) + 5 ( y + 3 )
= ( y+ 5)( y + 3)
Thay vào
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)
x3 - 3x2 - 9x - 5 = (x3 - 5x2) + (2x2 -10x) + (x - 5) = x2 (x - 5) + 2x(x - 5) + (x - 5) = (x - 5)(x2 + 2x + 1) = (x - 5)(x + 1)2
x^3-3x^2-9x-5
=x^3-5x^2+2x^2-10x+x-5
=x^2(x-5)+2x(x-5)+(x-5)
=(x-5)(x^2+2x+1)
=(x-5)(x+1)^2
= [ 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)