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B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(\left(\frac{2x+1}{\sqrt{x}^3-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right).\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\)
\(=\left(\frac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\)
\(=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{1}{\sqrt{x}-1}.\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{1-\sqrt{x}+x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1-\sqrt{x}+x-x+\sqrt{x}}{\sqrt{x}-1}=\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{x-1}\)
\(C=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Ta co:
\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)
\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)
\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
a/ \(B=\frac{1+x}{1+\sqrt{x}+x}\)
b/ Giải phương trình bậc 2 thì dễ rồi ha
c/ \(\frac{1+x}{1+\sqrt{x}+x}>\frac{2}{3}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)đung vì x khac 1
Phương trình bậc hai là\(x-\sqrt{6x}+1=0\) thì giải làm sao bạn ơi??
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
Ta có
\(1P=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{x\sqrt{x}-1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\frac{1}{\sqrt{x}-1}.\frac{x\sqrt{X}-x-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=1\frac{x\sqrt{x}-x-\sqrt{x}-1}{x-1}\)
Ta có thao câu b thì 1 - x > 0
<=> x < 1
=> \(0\le x< 1\)
Ta có \(P\sqrt{1-x}=\frac{x\sqrt{x}-x-\sqrt{x}-1}{-\sqrt{1-x}}< 0\)
\(\Leftrightarrow x\sqrt{x}-x-\sqrt{x}-1>0\)
Ta thấy \(0\le x< 1\Rightarrow x\sqrt{x}< x+\sqrt{x}+1\)
Vậy không có giá trị nào của x để cái trên xảy ra
\(P=\left(\frac{x\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
a) \(P=\left[\frac{x\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\left[\frac{x\sqrt{x}}{x\sqrt{x}-1}-\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{x\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{x\sqrt{x}-1}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=\frac{-x-\sqrt{x}}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{-\sqrt{x}}{\sqrt{x}-1}\)
vậy \(P=-\frac{\sqrt{x}}{\sqrt{x}-1}\) với \(x\ge0;x\ne1\)
b) để \(P>1\Leftrightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{-2\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\hept{\begin{cases}-2\sqrt{x}+1>0\\\sqrt{x}-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}-2\sqrt{x}+1< 0\\\sqrt{x}-1< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}< \frac{1}{2}\\\sqrt{x}>1\end{cases}}\) hoặc \(\hept{\begin{cases}\sqrt{x}>\frac{1}{2}\\\sqrt{x}< 1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< \frac{1}{4}\\x>1\end{cases}\left(loai\right)}\) hoặc \(\hept{\begin{cases}x>\frac{1}{4}\\x< 1\end{cases}}\)
\(\Rightarrow\frac{1}{4}< x< 1\)
kết hợp với \(ĐKXĐ:x\ge0;x\ne1\) thì ta có \(\frac{1}{4}< x< 1\)