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a) Với \(x\ge0;x\ne1\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-\left(3x-9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Vậy : \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b) \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
\(A_{max}\Leftrightarrow\left(\frac{17}{\sqrt{x}+3}\right)_{max}\)
Vì \(x\ge0;x\ne1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\frac{17}{\sqrt{x}+3}>0\end{cases}A_{max}\Leftrightarrow}\left(\sqrt{x}+3\right)_{min}\Leftrightarrow\sqrt{x}_{min}\Leftrightarrow x=0\)
Vậy : \(A_{max}=\frac{17}{3}\Leftrightarrow x=0\)
c,d chưa làm được .-.
c) Để \(A=\frac{1}{2}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{1}{2}\)
<=> \(-10\sqrt{x}+4=\sqrt{x}+3\)
<=> \(-11\sqrt{x}=-1\)
<=> \(\sqrt{x}=\frac{1}{11}\)
<=> \(x=\frac{1}{121}\left(tm\right)\)
Vậy ...
d) \(A\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\le0\)
<=> \(\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\sqrt{x}+9}\le0\)
<=> \(\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)
Vì \(\hept{\begin{cases}-17\sqrt{x}\le0\\3\sqrt{x}+9>0\end{cases}}\) \(\Rightarrow\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)(luôn đúng)
=> Ta có ĐPCM
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
\(A=\)\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\) \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(-\frac{\sqrt{x}+x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+x+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
= \(\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+x+1}\)
học tốt
\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(A=\frac{x+2}{\sqrt{x}^3-1^3}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có : x + 1 \(\ge\)\(2\sqrt{x}\)nên \(x+\sqrt{x}+1\ge3\sqrt{x}\)
\(\Rightarrow A=\frac{\sqrt{x}}{x+\sqrt{x}+1}\le\frac{\sqrt{x}}{3\sqrt{x}}=\frac{1}{3}\)
Vậy GTLN của A là \(\frac{1}{3}\)\(\Leftrightarrow x=1\)
\(a,A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{3}{\sqrt{x}+3}\)
\(\Leftrightarrow\frac{-\left(15\sqrt{x}-11\right)-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-3\left(1-\sqrt{x}\right)}{-x-2\sqrt{x}+3}\)
\(\Leftrightarrow\frac{-15\sqrt{x}+11-3x-9\sqrt{x}+2\sqrt{x}+6-3+3\sqrt{x}}{-x-2\sqrt{x}+3}\)
\(\Leftrightarrow\frac{19\sqrt{x}+3x-14}{x+2\sqrt{x}-3}\)
\(b,\) Xét \(\frac{19\sqrt{x}+3x-14}{x+2\sqrt{x}-3}\) phân tử \(13\sqrt{x}-5\)
Vậy để biểu thức trên nguyên thì \(13\sqrt{x}-5=0\)
\(\Leftrightarrow x=\left(\frac{5}{13}\right)^2\)
Vậy .......................
~~~~~~~~~ Học Tốt ~~~~~~~~~
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 \(maxP=\frac{7}{4}\)\(\Leftrightarrow x=0\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a.
\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b.Ta co:
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\le-5+\frac{17}{3}=\frac{2}{3}\)
Dau '=' xay ra khi \(x=0\)
Vay \(A_{max}=\frac{2}{3}\)khi \(x=0\)