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a) ĐKXĐ: \(a>0;a\ne1\)
b) ta có:
\(P=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)\)
c) ta có:
\(P=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi : \(a=\frac{1}{4}\)
Vậy min P =-1/4 khi a=1/4
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\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}\left(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
\(=\frac{\left(a-1\right)\left(-2\right)2\sqrt{a}}{4a}=-\frac{\left(a-1\right)}{\sqrt{a}}\)
a.\(DK:x\ge0\)
\(A=\frac{x-2\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}+1}=\sqrt{x}+1\)
b.Dat \(P=\frac{1}{A}\left(x+3\right)=\frac{x+3}{\sqrt{x}+1}\left(P>0\right)\)
\(\Rightarrow P\sqrt{x}+P=x+3\)
\(\Leftrightarrow x-P\sqrt{x}+3-P=0\)
Dat \(t=\sqrt{x}\left(t\ge0\right)\)
Ta co:
\(\Delta\ge0\)
\(\Leftrightarrow P^2-4\left(3-P\right)\ge0\)
\(\Leftrightarrow P^2+4P-12\ge0\)
\(\Leftrightarrow\left(P-2\right)\left(P+6\right)\ge0\)
TH1:
\(\hept{\begin{cases}P-2\ge0\\P+6\ge0\end{cases}\Leftrightarrow P\ge2}\)
TH2:
\(\hept{\begin{cases}P-2\le0\\P+6\le0\end{cases}\Leftrightarrow P\le2\left(P>0\right)}\)
Vi la de bai tim min nen lay TH1 thoi
Dau '=' xay ra khi \(x=\frac{P}{2}=1\)
Vay \(P_{min}=2\)khi \(x=1\)
a) ĐKXĐ: \(a\ne1;a\ne0\))
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a+1}}=\frac{\sqrt{a}\left(a-1\right)}{\sqrt{a+1}}\)
a/ ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
\(A=\left[\frac{1}{\sqrt{x}-1}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}+4}{\sqrt{x}-1}\right]\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+1}\)
b/
Ta có: \(A=\frac{1}{\sqrt{x}+1}\ge1\)
Vậy Min A = 1 .Dấu "=" xảy ra khi x = 0
a , rút gọn : A= \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right):\left(2-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{1\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{\sqrt{x}+1+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+4}{\sqrt{x}-1}\right)\)
A= \(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
A = \(\frac{1}{\sqrt{x}+1}\)
a/ Điều kiện \(\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\) \(\Rightarrow0\le a\ne\frac{1}{9}\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(a-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(1-3\sqrt{a}\right)+\left(\sqrt{a}-2\right)\left(1+3\sqrt{a}\right)+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}:\left(\frac{3a\sqrt{a}-2\sqrt{a}+6-a}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}-6a+\sqrt{a}+3a-2-6\sqrt{a}+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}.\left(\frac{3\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\right)\)
\(=\frac{3a-2\sqrt{a}-1}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
Hình như đề sai rồi bạn :(
a/ Điều kiện xác định : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}\Leftrightarrow}0\le a\ne9\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(1-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(3\sqrt{a}-1\right)+\left(2-\sqrt{a}\right)\left(3\sqrt{a}+1\right)-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}:\frac{\sqrt{a}+5}{3\sqrt{a}-1}\)
\(=\frac{6a-2\sqrt{a}+6\sqrt{a}+2-3a-\sqrt{a}-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}.\frac{3\sqrt{a}-1}{\sqrt{a}+5}\)
\(=\frac{3a-2\sqrt{a}-1}{3\sqrt{a}+1}.\frac{1}{\sqrt{a}+5}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(3\sqrt{a}+1\right)\left(\sqrt{a}+5\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+5}\)
c/ \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) thay vào M được
\(\frac{\sqrt{5}-2-1}{\sqrt{5}-2+5}=\frac{\sqrt{5}-3}{\sqrt{5}+3}=\frac{-7+3\sqrt{5}}{2}\)
d/ \(M=\frac{\sqrt{a}-1}{\sqrt{a}+5}=\frac{\sqrt{a}+5-6}{\sqrt{a}+5}=1-\frac{6}{\sqrt{a}+5}\)
Với mọi \(0\le a\ne9\) thì ta luôn có \(\sqrt{a}+5\ge5\Leftrightarrow\frac{6}{\sqrt{a}+5}\le\frac{6}{5}\Leftrightarrow-\frac{6}{\sqrt{a}+5}\ge-\frac{6}{5}\Leftrightarrow1-\frac{6}{\sqrt{a}+5}\ge1-\frac{6}{5}\)
\(\Rightarrow M\ge-\frac{1}{5}\)
Đẳng thức xảy ra khi a = 0
Vậy giá trị nhỏ nhất của M bằng \(-\frac{1}{5}\) khi a = 0
Điều kiện xác định \(a>0,a\ne1.\)
Ta có \(A=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\frac{a}{\sqrt{a}+1}=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\frac{a}{\sqrt{a}+1}\)
\(=\frac{a-1}{\sqrt{a}}\cdot\frac{a}{\sqrt{a}+1}=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\frac{a}{\sqrt{a}+1}=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}.\)
Vậy \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}.\) Dấu bằng xảy ra khi \(\sqrt{a}=\frac{1}{2}\to a=\frac{1}{4}\). Vậy GTNN của \(A\) là \(\frac{1}{4}.\)
A= \(\frac{2019}{x-\sqrt{x}+1}\)
Tìm GTLN của A