Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)
Với x>1 thoả mãn yêu cầu.
a) \(E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(E=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(E=\frac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(E=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(E=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(E=\frac{\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)}\)
b)đkxđ: \(x\ne1\); x\(\ge0\)
E=\(\frac{1}{3}\)<=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{1}{3}\)
<=>3(-5\(\sqrt{x}\)+2)=\(\sqrt{x}+3\)
<=>-15\(\sqrt{x}+6\)\(-\sqrt{x}\)=3
<=>\(-16\sqrt{x}=-3\)
<=>\(\sqrt{x}=\frac{3}{16}\)
\(< =>\left\{{}\begin{matrix}x=\frac{9}{256}\left(tm\right)\\x=\frac{-9}{256}\left(ktm\right)\end{matrix}\right.\)
vậy S=\(\left\{\frac{9}{256}\right\}\)