Bài 4:
P = ($\sqrt{x}$ - $\frac{1}{\sqrt{x}}$) : ($\frac{\sqrt{x}-1}{\sqrt{x}}$ + $\frac{1-\sqrt{x}}{x+\sqrt{x}}$)
ĐKXĐ: x > 0, x $\neq$ 1
P = $\frac{x-1}{\sqrt{x}}$ : [$\frac{(\sqrt{x}-1).(\sqrt{x}+1)}{\sqrt{x}.(\sqrt{x}+1)}$ + $\frac{1-\sqrt{x}}{\sqrt{x}.(\sqrt{x}+1)}$]
P = $\frac{x-1}{\sqrt{x}}$ : ($\frac{x-1+1-\sqrt{x}}{\sqrt{x}.(\sqrt{x}+1)}$)
P = $\frac{x-1}{\sqrt{x}}$ . ($\frac{\sqrt{x}.(\sqrt{x}+1)}{x-\sqrt{x}}$)
P = $\frac{(\sqrt{x}+1)^2}{\sqrt{x}}$
b) x = $\frac{2}{2+\sqrt{3}}$ ⇔ x = 4 - 2$\sqrt{3}$
Thay x = 4 - 2$\sqrt{3}$ vào P ta có:
P = $\frac{(\sqrt{4 - 2\sqrt{3}}+1)^2}{\sqrt{4 - 2\sqrt{3}}}$
P = $\frac{3}{\sqrt{(\sqrt{3}-1)^2}}$
P = $\frac{3}{\sqrt{3}-1}$
P = $\frac{3+3\sqrt{3}}{2}$
Vậy khi x = $\frac{2}{2+\sqrt{3}}$ thì P = $\frac{3+3\sqrt{3}}{2}$
c)
Bài 6:
A =$\frac{2\sqrt{x}}{\sqrt{x}-3}$ - $\frac{x+9\sqrt{x}}{x-9}$ và B = $\frac{x+5\sqrt{x}}{x-25}$ (với x $\geq$ 0; x $\neq$ 9; x $\neq$ 25)
a) A =$\frac{2\sqrt{x}}{\sqrt{x}-3}$ - $\frac{x+9\sqrt{x}}{x-9}$
A = $\frac{2\sqrt{x}.(\sqrt{x}+3)}{(\sqrt{x}-3).(\sqrt{x}+3)}$ - $\frac{x+9\sqrt{x}}{(\sqrt{x}-3).(\sqrt{x}+3)}$
A = $\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{(\sqrt{x}-3).(\sqrt{x}+3)}$
A = $\frac{x-3\sqrt{x}}{(\sqrt{x}-3).(\sqrt{x}+3)}$
A = $\frac{x-3\sqrt{x}}{(\sqrt{x}-3).(\sqrt{x}+3)}$
A = $\frac{\sqrt{x}}{\sqrt{x}+3}$
B = $\frac{x+5\sqrt{x}}{x-25}$
B = $\frac{\sqrt{x}}{\sqrt{x}-5}$
b) P = A : B
P = $\frac{\sqrt{x}}{\sqrt{x}+3}$ : $\frac{\sqrt{x}}{\sqrt{x}-5}$
P = $\frac{\sqrt{x}}{\sqrt{x}+3}$ . $\frac{\sqrt{x}-5}{\sqrt{x}}$
P = $\frac{\sqrt{x}-5}{\sqrt{x}+3}$
Xét hiệu P - 1, ta có:
P - 1 = $\frac{\sqrt{x}-5}{\sqrt{x}+3}$ - 1
P - 1 = $\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}$
P - 1 = $\frac{-8}{\sqrt{x}+3}$
Xét tử ta có: -8<0 (luôn đúng) ⇒ Tử âm
Xét mẫu ta có: $\sqrt{x}+3$ $\geq$ 3 ⇒ Mẫu dương
⇒ P - 1 < 0 hay P < 1
c) P = $\frac{\sqrt{x}-5}{\sqrt{x}+3}$
Ta có: $\sqrt{x}+3$ $\geq$ 3
⇔ $\frac{1}{\sqrt{x}+3}$ $\geq$ $\frac{1}{3}$
⇔ $\frac{-8}{\sqrt{x}+3}$ $\leq$ $\frac{-8}{3}$
⇒ P min = $\frac{-8}{3}$
Để P min ⇔ $\sqrt{x}+3$ = 3 hay x = 0 (TMĐK)
Vậy P min $\frac{-8}{3}$ khi x = 0
Bài 7:
M = ($\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}$ - $\frac{\sqrt{x}-2}{x-1}$) và N = $\frac{\sqrt{x}+1}{\sqrt{x}}$
với x > 0, x $\neq$ 1
Ta có: M = [$\frac{(\sqrt{x}+2).(\sqrt{x}-1)}{(\sqrt{x}+1)^2.(\sqrt{x}-1)}$ - $\frac{(\sqrt{x}-2).(\sqrt{x}+1)}{(\sqrt{x}+1)^2.(\sqrt{x}-1)}$]
M = $\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{(\sqrt{x}+1)^2.(\sqrt{x}-1)}$
M = $\frac{2\sqrt{x}}{(\sqrt{x}+1)^2.(\sqrt{x}-1)}$
a) A = M.N
A = $\frac{2\sqrt{x}}{(\sqrt{x}+1)^2.(\sqrt{x}-1)}$.$\frac{\sqrt{x}+1}{\sqrt{x}}$
A = $\frac{2}{x-1}$
b) A < -1
⇔ $\frac{2}{x-1}$ < -1
⇔ $\frac{2}{x-1}$ + 1 < 0
⇔ $\frac{2+x-1}{x-1}$ < 0
⇔ $\frac{x+1}{x-1}$ < 0
TH1: $\left \{ {{x+1<0} \atop {x-1>0}} \right.$
⇔ $\left \{ {{x<-1} \atop {x>1}} \right.$ ( vô lí)
TH2: $\left \{ {{x+1>0} \atop {x-1<0}} \right.$
⇔ $\left \{ {{x>-1} \atop {x<1}} \right.$
⇔ -1<x<1 kết hợp với ĐKXĐ
Vậy 0<x<1 thì A < -1
