## a. (x ³ + 4x +5a) . 4b b. (3x+5) . (3x ² + 3x + 7) c. (4x+7) . (x-3) + (3x+1) ² TÌM X a. (x-3) ² – x.(x+4)=0 b. x ³ – 4x = 0 c. 9x ³ – 16x = 0

Question

a. (x ³ + 4x +5a) . 4b
b. (3x+5) . (3x ² + 3x + 7)
c. (4x+7) . (x-3) + (3x+1) ²
TÌM X
a. (x-3) ² – x.(x+4)=0
b. x ³ – 4x = 0
c. 9x ³ – 16x = 0

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2 years 2020-11-21T01:59:50+00:00 1 Answers 49 views 0

## Answers ( )

1. a, 4b(x³+4x+5a)

= 4x³b+16xb+20ab

b, (3x+5)(3x²+3x+7)

= 3x(3x²+3x+7)+5(3x²+3x+7)

= 9x³+9x²+21x+15x²+15x+35

= 9x³+24x²+36x+35

c, (4x+7)(x-3)+(3x+1)²

= x(4x+7)-3(4x+7)+[(3x)²+2.3x.1+1²]

= 4x²+7x-12x-21+9x²+6x+1

= 13x²+x-20

Tìm x

a, (x-3)²-x(x+4)=0

⇔ (x²-2x.3+3²)-x²-4x=0

⇔ x²-6x+9-x²-4x=0

⇔ 9-10x=0

⇔ -10x=-9

⇔ x=$\frac{9}{10}$

b, x³-4x=0

⇔ x(x²-4)=0

⇔ x(x²-2²)=0

⇔ x(x-2)(x+2)=0

⇔ x=0 hoặc x-2=0 hoặc x+2=0

⇔ x=0 hoặc x=2 hoặc x=-2

c, 9x³-16x=0

⇔ x(9x²-16)=0

⇔ x[(3x)²-4²]=0

⇔ x(3x-4)(3x+4)=0

⇔ x=0 hoặc 3x-4=0 hoặc 3x+4=0

⇔ x=0 hoặc 3x=4 hoặc 3x=-4

⇔ x=0 hoặc x=$\frac{4}{3}$ hoặc x=$\frac{-4}{3}$