sử dụng phương pháp miền giá trị

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Ta có: \(P=\frac{\sqrt{x}-4}{\sqrt{x}}\times\frac{x+\sqrt{x}+1}{\sqrt{x}-4}\)

 \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)\(\left(ĐK:x>0\right)\)

Ta lấy \(P-2=\frac{x+\sqrt{x}+1}{\sqrt{x}}-2\)

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

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

                       \(=\frac{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{3}{4}}{\sqrt{x}}\)

                      \(=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}\)

Vì \(x>0\Rightarrow\sqrt{x}>0\)

 \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}>0\)

\(\Rightarrow P-2>0\)

\(\Rightarrow P>2\)

Học tốt 

Rút gọn : \(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\) (ĐKXĐ : \(0\le x\ne1\) )

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

Vậy B < 1

Mau la \(\sqrt{X - 3} \) that sao

ace nào giải giúp với ạ

a) \(ĐKXĐ:\) \(x\ne1,x>0\)

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

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

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

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

Vậy \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\left(x\ne1,x>0\right)\)

b) Xét hiệu \(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

\(=\frac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\forall x>0,x\ne1\)

Do đó : \(P>3\)

a. ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

b. Ta có \(B-5=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}-5=\frac{2x-3\sqrt{x}+2}{\sqrt{x}}=\frac{2\left(x-2.\sqrt{x}.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{8}+2}{\sqrt{x}}\)

\(=\frac{2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}}{\sqrt{x}}\)

Ta thấy \(\hept{\begin{cases}2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}>0\\\sqrt{x}>0\forall x>0\end{cases}\Rightarrow B-5>0\Rightarrow B>5}\)

Vậy \(B>5\)

B= 0,5 <=> \(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=0,5\)

<=> \(2.\left(2-5\sqrt{x}\right)=\sqrt{x}+3\) <=> 4 - 10\(\sqrt{x}\) = \(\sqrt{x}\) + 3

<=> 11\(\sqrt{x}\) = 1 <=> x = \(\frac{1}{11^2}=\frac{1}{121}\)(thỏa mãn)

c) Xét hiệu: B - \(\frac{2}{3}\) =   \(\frac{2-5\sqrt{x}}{\sqrt{x}+3}-\frac{2}{3}=\frac{6-15\sqrt{x}-2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\) Với mọi x > = 0 

=> \(B\le\frac{2}{3}\)

Giúp mình đi mình rút gọn đi đi lại lại mà chẳng ra

ĐK: x > = 0; x \(\ne\)1

\(B=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

= \(\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-1\right)}\)\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-1\right)}=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-1\right)}=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-1\right)}\)

= \(\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-1\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)