Giải phương trình :
[tex]\sqrt{3x+1}+\sqrt{2x-1}=4-x^{3}[/tex]
Đkxđ: [tex]\sqrt{3x+1}\geq 0[/tex] => [tex]x\geq \frac{-1}{3}[/tex]
[tex]\sqrt{2x-1}\geq 0=>x\geq \frac{1}{2}[/tex]
[tex]\sqrt{3x+1}+\sqrt{2x-1}=4-x^{3}[/tex]
[tex]=> (\sqrt{3x+1}-2)+(\sqrt{2x-1}-1)=1-x^3[/tex]
[tex]<=> \frac{3x+1-4}{\sqrt{3x+1}+2}+\frac{2x-1-1}{\sqrt{2x-1}+1}=1-x^3[/tex]
[tex]<=> \frac{3x-3}{\sqrt{3x+1}+2}+\frac{2x-2}{\sqrt{2x-1}+1}=1-x^3[/tex]
[tex]=> (x-1)(\frac{1}{\sqrt{3x+1}}+\frac{1}{\sqrt{2x-1}+1}+1+x^2+x)=0[/tex]
Vì vế sau >0 với mọi x
=> [tex]x-1=0=>x=1[/tex](TM ĐKXĐ)
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