Lời giải:
ĐKXĐ: $x>0$
Áp dụng BĐT Cô-si:
$x+2\geq 2\sqrt{2x}$
$\Rightarrow x+2\sqrt{x}+2\geq \sqrt{x}(2+2\sqrt{2})$
$\Rightarrow \frac{x+2\sqrt{x}+2}{\sqrt{x}}\geq 2+2\sqrt{2}$
$\Rightarrow P=-\frac{x+2\sqrt{x}+2}{\sqrt{x}}\leq -(2+2\sqrt{2})$
Vậy $P_{\max}=-(2+2\sqrt{2})$ khi $x=2$
rút gọn
B=\(\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)tìm đk để B rút gọn
C=\(\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)tìm x ∈Z để C ∈Z
Cho M= \(\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
a) Rút gọn M
b) Tìm các giá trị của x để có \(\dfrac{5}{3}M\) = \(\sqrt{x}+4\)
Tìm Min và Max(nếu có)
A=2x-\(\sqrt{x}\)
B=x+\(\sqrt{x}\)
C=1+\(\sqrt{2-x}\)
D=\(\sqrt{-x^2+2x+5}\)
E=\(\dfrac{1}{2x-\sqrt{x}+3}\)
F=\(\dfrac{1}{3-\sqrt{1-x^2}}\)
P=\(\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
Rút gọn
a. \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+x}{x-\sqrt{x}+1}+x+1\) với x > 0
b. \(\left(\dfrac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(\dfrac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\right)\) với x > 0
a. \(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{1}{1+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
b/\(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}với\) x>=0
c.\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}với\) x>=0,y>=0 và x≠y
Bải 1 :Rút gọn :
\(M=\left(\dfrac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\)\(\left(\dfrac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\right)\)
Bài 2 : Rút gọn :
\(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\)\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
trục căn thức
a) \(\dfrac{1}{\sqrt{x-1}};\dfrac{a+2}{\sqrt{a^2-4}};\dfrac{x-y}{\sqrt{x^2-y^2}};\dfrac{a}{\sqrt{x^2}}\) (n lẻ)
b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}\)
c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}\)
Cm
a.\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
b. \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=-\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>o và x≠y
