Giúp mjk toán,khó

D

dien0709

[TEX]\left{\begin{x^3=3xy+3}\\{y^3=xy-2}\right\to \left{\begin{x=\frac{y^3+2}{y}}(*)\\{(\frac{y^3+2}{y})^3=3(y^3+2)+3(**)[/TEX]

[TEX]t=y^3+2\to y^3=t-2(**)\to t^3-3t^2+3t+6=0\to (t-1)^3=-7[/TEX]

[TEX]\to t=1+\sqrt[3]{-7}\to y^3=-1+\sqrt[3]{-7}\to y=...\to x=...[/TEX]
 
D

dien0709

[TEX]\left{\begin{\sqrt[]{x}+\sqrt[]{y}=3}\\{\sqrt[]{x+5}+\sqrt[]{y+3}\leq a}\right \to \left{\begin{y=x-6\sqrt[]{x}+9}\\{\sqrt[]{x+5}+\sqrt[]{x-6\sqrt[]{x}+12}\leq a\right \to 3\geq \sqrt[]{x}\to x\leq 9[/TEX]

[TEX]t=\sqrt[]{x}; 4\leq x\leq 9\to 2\leq t\leq 3\to f(t)=\sqrt[]{t^2+5}+\sqrt[]{t^2-6t+12}[/TEX]

[TEX]f'(t)=\frac{t}{\sqrt[]{t^2+5}}+\frac{t-3}{\sqrt[]{t^2-6t+12}};f'(t)=0\to t=\frac{15-\sqrt[]{135}}{2} [/TEX]

Lập bảng biến thiên =>f(t) tăng với [TEX]\right t\in [2;3]ycbt\to a\geq f(3)=\sqrt[]{14}+\sqrt[]{3}[/TEX]
 
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