Cho \(A=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x\sqrt{x}-x}{1-\sqrt{x}}\)
a)Rút gọn A
b)Tìm x để A>0
MONG CÁC BẠN ZẢI NHANH ZÚP
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
11 tháng 10 2017
ĐK: tự ghi nha
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\left(\frac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\left(\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\frac{3+\sqrt{x}-1}{\sqrt{x}-1}\)
\(P=\frac{3}{\sqrt{x}-1}+1\)
P/s : Ko biết có đúng ko
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
28 tháng 7 2017
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
NM
Nguyễn Minh Quang
Giáo viên
29 tháng 8 2021
ta có
\(A=\frac{\sqrt{x}-\sqrt{x-1}-\left(\sqrt{x}+\sqrt{x-1}\right)}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{0}{1-\sqrt{x}}\)
\(=-\frac{2\sqrt{x-1}}{x-\left(x-1\right)}=-2\sqrt{x-1}\) dễ thấy \(A\le0\) với mọi x
a. A=Đề=\(\frac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\frac{x\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\)\(\left(ĐKXĐ:x>1\right)\)
\(=\frac{-2\sqrt{x-1}}{x-x+1}+x\)\(=x-2\sqrt{x-1}\)
b. A>0 \(\Leftrightarrow x-2\sqrt{x-1}>0\)
\(\Leftrightarrow x>2\sqrt{x-1}\)\(\Rightarrow x^2>4\left(x-1\right)\)\(\Leftrightarrow x^2>4x-4\)
\(\Leftrightarrow x^2-4x+4>0\)\(\Leftrightarrow\left(x-2\right)^2>0\)\(\Rightarrow x-2>0\)
\(\Leftrightarrow x>2\)
a) A= \(\frac{1}{\sqrt{x}+\sqrt{ }x-1}\) - \(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x\sqrt{x}-x}{1-\sqrt{x}}\) với x>1\(\frac{\sqrt{x}-\sqrt{x-1}}{x-\left(x-1\right)}-\frac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\) \(=\frac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{1}+x\) \(=-2\sqrt{x-1}+x\) b) với x>1 ta có A>0 hay \(-2\sqrt{x-1}\)\(+x\)\(>0\)\(\Rightarrow x>2\sqrt{x-1}\)\(\Leftrightarrow\)\(x^2>4\left(x-1\right)\Leftrightarrow x^2-4x+4>0\)\(\left(x-2\right)^2>0\)(--> \(x\ne\pm2\) )