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Ta có : x2 - 6x + 5
= x2 - x - 5x + 5
= (x2 - x) - (5x - 5)
= x(x - 1) - 5(x - 1)
= (x - 5)(x - 1)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
ak
x8 + -7x4 + -8 = 0 Reorder the terms: -8 + -7x4 + x8 = 0 Solving -8 + -7x4 + x8 = 0 Solving for variable 'x'. Factor a trinomial. (-1 + -1x4)(8 + -1x4) = 0
2x4 - 3x3 - 7x2 +6x+8
= 2x4 - 4x3 + x3 - 2x2 - 5x2 +10x - 4x +8
= 2x3.(x-2) +x2.(x-2) - 5x.(x-2) - 4.(x-2)
= (x-2).(2x3 +x2 - 5x -4)
= (x-2).(2x3 + 2x2 - x2 - x - 4x-4)
= (x-2).(x+2).(2x2 -x -4)
....
x3 - 9x2 + 6x + 16
= x3 - 8x2 -x2 + 8x - 2x + 16
= x2(x-8) -x(x-8) -2(x-8)
= (x-8)(x2-x-2)
= (x-8)(x2-2x + x - 2)
=(x-8)[x(x-2)+(x-2)]
=(x-8)(x-2)(x+1)