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ĐK: x > 0
a) Rút gọn M
M = \(\frac{\sqrt{x}}{x+\sqrt{x}}:\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)
=> M \(\le\)1/3
=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn
Vậy max M = 1/3 tại x = 1
ĐK: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}-3\right)+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x-6\sqrt{x}+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-9x+9}{x-9}\)
Xét P-1 = \(\frac{\sqrt{x}+3}{\sqrt{x}+2}-1\)
P-1 = \(\frac{\sqrt{x}+3-\sqrt{x}-2}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+2}\)
Nhận xét : \(\hept{\begin{cases}1>0\\\sqrt{x}+2>0\end{cases}}vớimoix\)
-> P-1 >0 với mọi x
-> P>1
Thay x=6-2 căn 5 vào P -> P=\(\frac{\sqrt{6-2\sqrt{5}}+3}{\sqrt{6-2\sqrt{5}+2}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+3}\)
=\(\frac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\frac{\sqrt{5}+3}{\sqrt{5}+1}\)
\(P=\frac{\sqrt{x}+3}{\sqrt{x}+2}\)( ĐKXĐ : \(x\ge0\))
1) Ta có : \(P=\frac{\sqrt{x}+3}{\sqrt{x}+2}=\frac{\sqrt{x}+2+1}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)
Vì \(\frac{1}{\sqrt{x}+2}>0\left(\forall x\ge0\right)\)
Cộng 1 vào mỗi vế => \(1+\frac{1}{\sqrt{x}+2}>1\)
Vậy P > 1
2) Với \(x=6-2\sqrt{5}\)( tmđk )
Khi đó \(P=1+\frac{1}{\sqrt{6-2\sqrt{5}}+2}\)
\(P=1+\frac{1}{\sqrt{5-2\sqrt{5}+1}+2}\)
\(P=1+\frac{1}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}\)
\(P=1+\frac{1}{\left|\sqrt{5}-1\right|+2}\)
\(P=1+\frac{1}{\sqrt{5}-1+2}\)
\(P=1+\frac{1}{\sqrt{5}+1}\)
\(P=\frac{\sqrt{5}+1}{\sqrt{5}+1}+\frac{1}{\sqrt{5}+1}\)
\(P=\frac{\sqrt{5}+1+1}{\sqrt{5}+1}=\frac{\sqrt{5}+2}{\sqrt{5}+1}\)
Bạn tự thu gọn thành 1+\(\frac{1}{\sqrt{x}+2}\) <= 1+\(\frac{1}{2}\)=\(\frac{3}{2}\) <=> x = 0
a) Với \(x\ge0\)và \(x\ne1\)ta có:
\(P=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)
\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-2x+5\sqrt{x}-3-x-5\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-\left(3x-10\sqrt{x}+7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}\)
b) \(P=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}=\frac{-3\sqrt{x}-12+19}{\sqrt{x}+4}=\frac{-3\left(\sqrt{x}+4\right)+19}{\sqrt{x}+4}=-3+\frac{19}{\sqrt{x}+4}\)
Vì \(x\ge0\); \(x\ne1\)\(\Rightarrow\sqrt{x}+4\ge4\)
\(\Rightarrow\frac{19}{\sqrt{x}+4}\le\frac{19}{4}\)\(\Rightarrow P\le-3+\frac{19}{4}=\frac{7}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow x=0\)( thỏa mãn )
Vậy \(maxP=\frac{7}{4}\)\(\Leftrightarrow x=0\)
Giải:
ĐKXĐ của P là \(x\ge2\)và \(x\ne5\)
Phân tích tử:
x-5 = x-2-3
= (\(\sqrt{x-2}\)-\(\sqrt{3}\))(\(\sqrt{x-2}\)+\(\sqrt{3}\))
Xét P=\(\frac{\left(\sqrt{x-2}-\sqrt{3}\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{\sqrt{x-2}-\sqrt{3}}\)
= \(\sqrt{x-2}+\sqrt{3}\)
=> Min P= \(\sqrt{3}\)khi X=2.
Mình chỉ có thể tìm GTNN, còn GTLN thì mk chịu.