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\(ĐKXĐ:x\ge0\)
Đề sai???
Sửa lại
\(a,P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+2-x+\sqrt{x}-1+x-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(\Leftrightarrow P=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)\)
\(\Leftrightarrow P=\frac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\frac{x+1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow P=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\frac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)
b) Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Để \(P\le0\Leftrightarrow-\sqrt{x}+1\le0\)
\(\Leftrightarrow-\sqrt{x}\le-1\)
\(\Leftrightarrow\sqrt{x}\ge1\)
\(\Leftrightarrow x\ge1\)
Vì đkxđ : \(x\ne1\)
Vậy để \(P\le0\Leftrightarrow x>1\)
ĐKXĐ: \(x\ge0\)
a/ Đề \(=\left(\frac{1-\sqrt{x}^3}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left[\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right]\left[\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\left(1+2\sqrt{x}+x\right)\left(1-2\sqrt{x}+x\right)\)
\(=\left(1+\sqrt{x}\right)^2\left(1-\sqrt{x}\right)^2\)
\(=\left[\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\right]^2=\left(1-x\right)^2\)
b/ \(P< 7-4\sqrt{3}\Leftrightarrow\left(1-x\right)^2< 7-4\sqrt{3}\)
\(\Rightarrow\left(1-x\right)^2< \left(2-\sqrt{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-x< 2-\sqrt{3}\Rightarrow x>-1+\sqrt{3}\\1-x< \sqrt{3}-2\Rightarrow x>3-\sqrt{3}\end{cases}}\)
Vậy \(x>3-\sqrt{3}\)
a) \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2\sqrt{x}}\)
\(A=\left(\frac{x+2}{\sqrt{x^3}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}-1}{2\sqrt{x}}\)
\(A=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}=\frac{2\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: A-\(\frac{2}{3}\)= \(\frac{\sqrt{x}}{x+\sqrt[]{x}+1}-\frac{2}{3}\)=\(\frac{6\sqrt{x}-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)
=\(\frac{-2\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}\)=\(\frac{-2}{3}.\frac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\)<0
hay A\(-\frac{2}{3}\)<0
=>A<\(\frac{2}{3}\)
ĐK: \(x>0,x\ne4\)
1, \(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{\sqrt{x}^3}\)
\(A=\frac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{x}\)
\(A=\frac{2}{\sqrt{x}+2}.\frac{x-4}{x}\)
\(A=\frac{2\sqrt{x}-4}{x}\)
2, \(x=\left(\sqrt{3}+1\right)^2\)
Thay \(x=\left(\sqrt{3}+1\right)^2\):
\(A=\frac{2\sqrt{3}-2}{\left(\sqrt{3}+1\right)^2}\)
3, \(A\ge\frac{1}{4}\Rightarrow\)\(\frac{2\sqrt{x}-4}{x}-\frac{1}{4}\ge0\)
\(\Leftrightarrow\frac{8\sqrt{x}-16-x}{4x}\ge0\)
\(\Rightarrow x-8\sqrt{x}+16\le0\)
\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(TM\right)\)
Lời giải:
ĐKXĐ: \(x\geq 0; x\neq 1\)
Ta có:
\(A=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}-1}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)