ĐKXĐ: x \(\ge\)0; x khác 9 (1)

a) B = \(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

B = \(\frac{-\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{-4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{4}{3-\sqrt{x}}\)

b) B > A <=> \(\frac{4}{3-\sqrt{x}}>1\) <=> \(\frac{4}{3-\sqrt{x}}-1>0\)

<=> \(\frac{4-3+\sqrt{x}}{3-\sqrt{x}}>0\)

<=> \(\frac{\sqrt{x}+1}{3-\sqrt{x}}>0\)

Do \(\sqrt{x}+1>0\) => \(3-\sqrt{x}>0\) <=> \(\sqrt{x}< 3\)

<=> \(x< 9\)

Kết hợp với đk (1)

=> \(0\le x< 9\)

a: \(5x-20x^2=0\)

\(\Leftrightarrow5x\left(1-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

c: \(x\left(x-3\right)-5x+15=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)

a, Ta có : \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)

b , Ta có : \(5xy\sqrt{\frac{x^2}{y^6}}=5xy\frac{x}{y^3}=\frac{5x^2}{y^2}\)

c, Ta có : \(0,2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0,2x^3y^3.\frac{4}{x^2y^4}=\frac{0,8x}{y}\)

Mình nhầm tìm GTLN

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a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)

\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)

\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)

b) Với \(0< x< 1\)\(\Rightarrow0< \sqrt{x}< 1\)

\(\Rightarrow\sqrt{x}-1< 0\)

mà \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-1\right)< 0\)

\(\Rightarrow-\sqrt{x}.\left(\sqrt{x}-1\right)>0\)\(\Rightarrow P>0\)( đpcm )

c) \(P=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)

\(=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)

\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)( thỏa mãn ĐKXĐ )

Vậy \(maxP=\frac{1}{4}\)\(\Leftrightarrow x=\frac{1}{4}\)

ĐKXĐ \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)

a,  Ta có \(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)

               \(P=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)

              \(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)

             \(P=\frac{2\sqrt{x}-2x}{\sqrt{2}}\)

             \(P=\sqrt{2x}-\sqrt{2}x\)

             \(P=\sqrt{2x}\left(1-\sqrt{x}\right)\)

b,        Vì \(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow1-\sqrt{x}< 1\)

                 \(\Rightarrow\sqrt{2x}\left(1-\sqrt{x}\right)>0\)

 c,        Ta có \(P=-\sqrt{2}\left(x-\sqrt{x}\right)\)  

                      \(P=-\sqrt{2}\left(x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)

                      \(P=-\sqrt{2x}\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{\sqrt{8}}\le\frac{1}{\sqrt{8}}\)

               Dấu = xảy ra \(\Leftrightarrow\)\(\sqrt{x}-\frac{1}{2}=0\)

                                      \(\Rightarrow x=\frac{1}{4}\)

             vậy GTLN của P là \(\frac{1}{\sqrt{8}}\)với x=\(\frac{1}{4}\)

Bài làm:

Ta có: 

\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)

\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)

\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)