Question

giải phương trình x√(3x-2)+√(3-2x)=√(x^3+x^2+x+1)

Answer 1

<=> x^2(3x-2)+2x(3x-2)+(3x-2)= x^3+x^2+x+1

<=>2x^3+3x^2-7x+2=0

<=>(2x^3-2x^2)+(5x^2-5x)+(-2x+2)+0

<=>(x-2)(2x^2+5x-2)=0

TH1 x-2=0 => x=2

TH2 2x^2+5x-2=0 <=>. (x căn2)^2+2*xcăn2*5/4+25/8-41/8=0

<=>(xcăn2+5/(2căn2))^2-(căn41/(2căn2))^2=0

(xcăn2+5/(2căn2)+3/(2căn2))*(xcăn2+5/(2căn2)-căn41/(2căn2))=0

Th2a (xcăn2+5/(2căn2)+căn41/(2căn2))=0

<=> x=(-5-căn41)/4

Th2b (xcăn2+5/(2căn2)-căn41/(2căn2))=0

<=> x=(-5+căn41)/4

TH2b (xcăn2+5/(2căn2)-3/(2căn2))=0

\<=>x=-1/2

Answer 2

nhầm TH2b =(-5+căn41)/4

Answer 3

\(x\sqrt{\left(3x-2\right)}+\sqrt{\left(3-2x\right)}=\sqrt{\left(x^3+x^2+x+1\right)}\)

\(\Leftrightarrow\left(x\sqrt{3x-2}+\sqrt{3-2x}\right)^2=\left(\sqrt{x^3+x^2+x+1}\right)^2\)

\(\Leftrightarrow3x^2-2x^2+2x\sqrt{3x-2}.\sqrt{3-2x}.x+3-2x=x^3+x^2+x+1\)

\(\Leftrightarrow2\sqrt{3x-2}.\sqrt{3-2x}.x=-2x^3+3x^2+3x+2\)

\(\Leftrightarrow\left(2\sqrt{3x-2}.\sqrt{3-2x}.x\right)^2=\left(-2x^3+3x^2+3x-2\right)\)

\(\Leftrightarrow-24x^4+52x^3-24x^2=4x^6-12x^5-3x^4+26x^3-3x^2-12x+4\)

\(\Rightarrow x=1\)