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1) $x^3$ - $3x^2$ + 2$\sqrt[2]{(x+2)^3}$ = 6x
[TEX]pt=>x^3-3x(x+2)+2\sqrt[]{(x+2)^3}=0[/TEX]
[TEX]t=\sqrt[]{x+2}\geq 0=>x^3-3xt^2+2t^3=0=>3t^3-t^3+x^3-3xt^2=0[/TEX]
[TEX]=>(x-t)(-2t^2+x^2+xt)=0=>(x-t)^2(x+2t)=0[/TEX]
[TEX]x=t\geq 0=>x=-1(l); x=2[/TEX]
[TEX]x=-2t\leq 0=>x=2+2\sqrt[]{3}(l); x=2-2\sqrt[]{3}[/TEX]
2. đk
đặt $\sqrt[]{2x-7}=a^2 \longrightarrow \sqrt[4]{2x-7}=a \longrightarrow x=\dfrac{a^4+7}{2} $
$\sqrt[]{x-3}=b^2 \longrightarrow \sqrt[4]{x-3}=b \longrightarrow x=b^4+3\\
pt \longleftrightarrow b^4+3+a^4+7+\dfrac{1}{a^2}+\dfrac{1}{b^2}=2(a+b+5)\\
\longleftrightarrow (\dfrac{1}{b^2}-2b+b^4)+(\dfrac{1}{a^2}-2a+a^4)=0\\
\longleftrightarrow (\dfrac{1}{b}-b^2)^2+(\dfrac{1}{a}-a^2)^2=0$
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