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[05479865132]

Giải phương trình

$4+2\sqrt{1-x}+3x=5\sqrt{x+1}+\sqrt{1-x^2 }$

$2\sqrt{4x^2-14x+16}+1=x+\sqrt{x^2-4x+5}$

$13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}=16$

$x+\sqrt{1+3x}+\sqrt{1-x}=\sqrt{1+2x}+\sqrt{1-2x}$

$x^{3}+\frac{\sqrt{68}}{x^{3}}=\frac{15}{x}$

x=$\sqrt[3]{\sqrt{1+x}+\sqrt{2+3x}}$

Giúp mình mấy bài này

Edited by 05479865132, 27-01-2017 - 07:18.

[Zeref]

3.$13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}=16$

Ta có $13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}=x(\sqrt{13}\sqrt{13-13x^2}+3\sqrt{3}\sqrt{3+3x^2}) \leq x\sqrt{40(16-10x^2)}=2\sqrt{10}x\sqrt{16-10x^2} \leq 16$