Đáp án:
a. $A = \frac{13}{12}$
b. $M = \frac{4\sqrt[]{x}+6}{\sqrt[]{x}+1}$
c. $M ≤ 6$
Giải thích các bước giải: ĐKXĐ : $x ≥ 0 ; x \ne 1$
a, $A = \frac{3}{\sqrt[]{x}-1} + \frac{\sqrt[]{x}+3}{x-1}$
$A = \frac{3×(\sqrt[]{x}+1)}{(\sqrt[]{x}-1)×(\sqrt[]{x}+1)} + \frac{\sqrt[]{x}+3}{(\sqrt[]{x}-1)×(\sqrt[]{x}+1)}$
$A = \frac{4\sqrt[]{x}+6}{(\sqrt[]{x}-1)×(\sqrt[]{x}+1)}$
Thay $x = 25$ vao A
$A = \frac{4\sqrt[]{25}+6}{((\sqrt[]{25}-1)×(\sqrt[]{25}+1)}$
$A = \frac{26}{24}$
$A = \frac{13}{12}$
b, $B = \frac{x+2}{x+\sqrt[]{x}-2} - \frac{\sqrt[]{x}}{\sqrt[]{x}+2}$
$B = \frac{x+2}{(\sqrt[]{x}-1)×(\sqrt[]{x}+2)} - \frac{\sqrt[]{x}×(\sqrt[]{x}-1)}{(\sqrt[]{x}+2)×(\sqrt[]{x}-1)}$
$B = \frac{\sqrt[]{x}+2}{(\sqrt[]{x}+2)×(\sqrt[]{x}-1)}$
$B = \frac{1}{\sqrt[]{x}-1}$
$M = A : B$
⇔ $M = \frac{4\sqrt[]{x}+6}{(\sqrt[]{x}-1)×(\sqrt[]{x}+1)} : \frac{1}{\sqrt[]{x}-1}$
⇔ $M = \frac{4\sqrt[]{x}+6}{(\sqrt[]{x}-1)×(\sqrt[]{x}+1)}×( \sqrt[]{x} - 1 )$
⇔ $M = \frac{4\sqrt[]{x}+6}{\sqrt[]{x}+1}$
c. $M = \frac{4\sqrt[]{x}+6}{\sqrt[]{x}+1}$
⇔ $M = \frac{4×(\sqrt[]{x}+1)+2}{\sqrt[]{x}+1}$
⇔ $M = 4 + \frac{2}{\sqrt[]{x}+1}$
Vì $\sqrt[]{x} + 1 ≥ 1 ∀ x ≥ 0 ; x \ne 1$
⇒ $\frac{2}{\sqrt[]{x}+1} ≤ \frac{2}{1} = 2$
⇒ $M = 4 + \frac{2}{\sqrt[]{x}+1} ≤ 4 + 2$
⇔ $M ≤ 6$
Dấu "=" xảy ra ⇔ $x = 0$
