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\(1.\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=|\sqrt{x-1}-1|=\sqrt{x-1}-1\)
\(2.\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(2x-3\right)^2}}=\dfrac{1}{|2x-3|}\)
\(3.\dfrac{1}{\sqrt{x+2\sqrt{x-1}}}=\dfrac{1}{\sqrt{x-1+2\sqrt{x-1}+1}}=\dfrac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}=\dfrac{1}{|\sqrt{x-1}+1|}\)
bạn cho mình hỏi tại sao từ \(\sqrt{x-1-2\sqrt{x-1}+1}\) sang \(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
ĐKXĐ :x\(\ge\)0
a) với x=64 thỏa mãn đk; khi đó: A=\(\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
b)với đk của x thì B xác định ; ta có
B\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
c)Xét M=A:B =\(\dfrac{2+\sqrt{x}}{\sqrt{2}}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Để \(M>\dfrac{3}{2}hay\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\Leftrightarrow2\sqrt{x}+4>3\sqrt{x}+3\left(do:\sqrt{x}+1>0\right)\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)
Kết hợp đk x\(\ge\)0. Vậy 0\(\le\)x<1 thì M=A:B>3/2
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-\left(x+4\right)}=\dfrac{4\sqrt{x}}{x+4}\)
b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)
a) ĐK ko bt đúng or sai
ĐKXĐ : \(\left\{{}\begin{matrix}x>=0\\x\ne0\\x\ne1\end{matrix}\right.\)
P =( \(\dfrac{1}{\left(\sqrt{x}\right)^2-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\)) . \(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
=\(\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
=\(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\sqrt{x}-1\)
\(\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\sqrt{x}-1\)
=\(\dfrac{-\sqrt{x}+1}{\sqrt{x}}\)
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{x-2\sqrt{x}+1}=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b: Khi x=9 thì \(A=\dfrac{-1}{3\left(3-1\right)}=\dfrac{-1}{6}\)
Bài 3:
a: \(A=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: \(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)+\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x+2-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}-1}=\sqrt{x}+1\)
\(Q=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\sqrt{x}+1\)
2.
\(x-2\sqrt{x}=\sqrt{x}(\sqrt{x}-3)+\frac{1}{4}(\sqrt{x}-3)+\frac{3}{4}(\sqrt{x}+1)\)
\(\geq \frac{3}{4}(\sqrt{x}+1)\)
\(\Rightarrow I\leq \frac{\sqrt{x}+1}{\frac{3}{4}(\sqrt{x}+1)}=\frac{4}{3}\)
Vậy $I_{\max}=\frac{4}{3}$ tại $x=9$
1. Với $x\geq \frac{1}{2}$ thì:
\(3x+\sqrt{x}+1=(\sqrt{2x}-1)(\sqrt{\frac{9}{2}x}-1)+(1+\frac{5\sqrt{2}}{2})\sqrt{x}\)
\(\geq (1+\frac{5\sqrt{2}}{2})\sqrt{x}\)
\(\Rightarrow H=\frac{\sqrt{x}}{3x+\sqrt{x}+1}\leq \frac{\sqrt{x}}{(1+\frac{5\sqrt{2}}{2})\sqrt{x}}=\frac{1}{1+\frac{5\sqrt{2}}{2}}=\frac{5\sqrt{2}-2}{23}\)
Đây chính là $H_{\max}$. Giá trị này đạt tại $x=\frac{1}{2}$