Question

[Min]

Giaỉ các PT:
a) \(\left(x^2+1\right)^2=4\left(2x-1\right)\)

b) x(x+1)(x+2)(x+3)=24

c)\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

d)\(\left(x^2-9\right)^2=12x+1\)

e)\(\left(x-1\right)^5+\left(x+3\right)^5=242\left(x+1\right)\)

Giup mk vs mn

Answer 1

[Mai Chi]

b) x(x+1)(x+2)(x+3)=24

(x²+3x)(x²+3x+2)=24 (1)

Đặt x²+3x+1=a

Khi đó(1)<=>(a-1)(a+1)=24

a²-1=24 <=>a²=25<=>a=5;-5

a=5=>x²+3x+1=5=>x²+3x-4=0=>(x-1)(x+4)=0=>x=1,-4

a=-5=>x^2+3x+1=-5=>x²+3x+6=0=>(x+\(\frac{3}{2}\))²+\(\frac{15}{4}\)=0

=>pt vô nghiệm