leminhnghiatt

CT này không khó. Thay $y=1-x$

$\frac{2^{2y+1}}{2^{2y}-2}$

$=\frac{2^{3-2x}}{2^{2-2x}-2}$

$=\frac{2^3.2^{-2x}}{2^{-2x}(2^2-2^{2x+1})}$

$=\frac{2^3}{2^2-2^{2x+1}}$

$=\frac{4}{2-2^{2x}}$

$=-\frac{4}{2^{2x}-2}$

$=> f(x)+f(y)=$$\frac{2^{2x+1}}{2^{2x}-2}$$-$$\frac{4}{2^{2x}-2}$$=$$\frac{2(2^{2x}-2)}{2^{2x}-2}$$=2$

ĐK: $x \not = \frac{1}{2}$

Xét:$ f(x)+f(y)$ với $x+y=1 => y=1-x$. Thay vào $f(y)$ Sau đó CM:

$f(x)+f(y)=\frac{2^{2x+1}}{2^{2x}-2}+\frac{2^{2y+1}}{2^{2y}-2}=2$

Khi đó áp dụng CT ta có: 

$A=1997+[f(\frac{1}{1998})+f(\frac{1997}{1998})]+...+[f(\frac{998}{1998})+f(\frac{1000}{1998})]$

$A=1997+2.998$

$A=3993$

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

$=> x^3+2+3x-2=3x-2+3\sqrt[3]{3x-2}$

$=> x^3+3x=(3x-2)+\sqrt[3]{3x-2}$

Đặt $\sqrt[3]{3x-2}=a$ thay vào ta có:

$x^3+3x=a^3+3a$

$=> (x-a)(x^2+xa+a^2-3)=0$

$=> \begin{cases} & \text{ } x=a \\ & \text{ } x^2+xa+a^2-3=0 \end{cases}$

Theo BĐT Bu-nhi-a:

$x\sqrt{1-y^2}+\sqrt{1-x^2}.y \leqslant \sqrt{(x^2+1-x^2)(1-y^2+y^2)}=1$

Dấu = xảy ra $=>\frac{x}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{y}$

$=> x^2+y^2=1$ 

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

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

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