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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)Ta có vế phải trái\(=x^2+x\sqrt{3}+1=x^2+2.\dfrac{\sqrt{3}}{2}x+\left(\dfrac{\sqrt{3}}{2}\right)^2-\left(\dfrac{\sqrt{3}}{2}\right)^2+1\)
\(=\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\) =vế phải
b)Ta có \(x^2+x\sqrt{3}+1=\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Vậy giá trị nhỏ nhất của biểu thức \(x^2+x\sqrt{3}+1\) là \(\dfrac{1}{4}\) khi \(\left(x+\dfrac{\sqrt{3}}{2}\right)^2=0\Leftrightarrow x=-\dfrac{\sqrt{3}}{2}\)
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
b ) \(x-\sqrt{3x}+1=x-2\cdot\frac{\sqrt{3}}{2}+\frac{3}{4}-\frac{3}{4}+1\)
\(=\left(\sqrt{x}-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\)
vì \(\left(\sqrt{x}-\frac{\sqrt{3}}{2}\right)^2\ge0\)với mọi x
=> \(\left(\sqrt{x}-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)voi moi x
=>\(\frac{1}{\left(\sqrt{x}-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}}\le\frac{1}{\frac{1}{4}}\le4\)
=> max A \(\le4\)
dau = xay ra <=> \(\left(\sqrt{x}-\frac{\sqrt{3}}{2}\right)=0\Leftrightarrow x=\frac{3}{4}\)
a) Với \(x>0;x\ne1\), ta có:
\(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Rightarrow2P=\frac{2\sqrt{x}+2}{\sqrt{x}}\)
\(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\left(ĐKXĐ:x\ne0\right)\left(1\right)\)
Mà theo đề bài : \(x>0\)nên phương trình luôn được xác định.
\(\left(1\right)\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}+5\right)}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=\sqrt{x}\left(2\sqrt{x}+5\right)\)
\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}+2-2x-5\sqrt{x}\)
\(\Leftrightarrow-2x-3\sqrt{x}+2=0\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}=1\\\sqrt{x}=-2\left(vn\right)\end{cases}}\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(TMĐK:x>0;x\ne1\right)\)
Vậy \(2P=2\sqrt{x}+5\Leftrightarrow x=\frac{1}{4}\)
a, Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định thì (x-1)(x-3)\(\ge\)0
TH1: \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\Leftrightarrow}x\ge3}\)TH2:\(\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\Leftrightarrow}x\le1}\) Vậy nếu \(x\ge3\) hoặc \(x\le1\) thì biểu thức có nghĩa
b, Để \(\sqrt{x^2-4}=\sqrt{\left(x-2\right)\left(x+2\right)}\)có nghĩa thì (x-2)(x+2)\(\ge0\)
TH1: \(\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge-2\end{matrix}\right.\Leftrightarrow x\ge}2}\)TH2:\(\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\le-2\end{matrix}\right.\Leftrightarrow}x\le-2}\)Vậy nếu \(x\ge2\) hoặc \(x\le-2\) thì biểu thức có nghĩa
\(a,Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}};x>0;x\ne1;x\ne4\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
\(a,\)\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{x-1}\)\(\left(đpcm\right)\)
\(b,Q=\frac{2}{x-1}\)
\(Q\in Z\Leftrightarrow\frac{2}{x-1}\in Z\Rightarrow x-1\inƯ_2\)
Mà \(Ư_2=\left\{\pm1;\pm2\right\}\)
TH1 : \(x-1=-1\Rightarrow x=0\)
TH2 : \(x-1=1\Rightarrow x=2\)
TH3 : \(x-1=-2\Rightarrow x=-1\)
TH4 :\(x-1=2\Rightarrow x=3\)
\(\Rightarrow\)x nguyên lớn nhất là 3 để Q là số nguyên