Question

giải pt:
a, 6x⁴-8x²+3=0
b,(x²-3x+3)(x²-2x+3)=2x²
c,(x²-5x+1)(x²-4)=6(x-1)²
d,x²+\(\frac{4x^2}{\left(x+2\right)^2}\)=12
e, \(\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\left(\frac{x^2-4}{x^2-1}\right)=0\)
f, \(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}-2=0\)
g, \(\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

Answer 1

a) x vô nghiệm

b)<=>(x²-3x+3)(x²-2x+3)-2x²=(x-3)(x-1)(x²-x+3)

=>(x-3)(x-1)(x²-x+3)=0

TH1:x-3=0

=>X=3

TH2:x-1=0

=>x=1

TH3:x²-x+3=0

<=>(-1)²-4(1.3)=-11

vì -11<0

=>x=1 hoặc 3

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