Question

\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)0

Giai phuong trinh

Answer 1

\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)

<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)

<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)

<=> \(3x^4+18x^2+12x-33=0\)

<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)

<=> \(x-1=0\)

<=> \(x=1\)

Mà vì: \(x^3+x^2+7x+11\ne0\)

=> x = 1

Answer 2

\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)

\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)

\(=>\left(x^2+3\right)^2=16\)

Đến đây tự tìm nha 

 Hok tốt