rút gọn \(\frac{\left(1-\frac{4}{\sqrt{x}+1}+\frac{1}{x-1}\right)}{\frac{x-2\sqrt{x}}{x-1}}\)
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
30 tháng 7 2019
a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)
\(M=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+2}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)
20 tháng 10 2023
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
10 tháng 2 2019
cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
29 tháng 7 2020
Trả lời:
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(A=\left[\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right]\div\left[\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\right]\)
\(A=\left[\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right]\div\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}+\frac{2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\right]\)
\(A=\left[\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right]\div\left[\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\right]\)
\(A=\left[\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right]\div\left[\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\right]\)
\(A=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\div\frac{1}{\sqrt{x}-1}\)
\(A=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\times\frac{\sqrt{x}-1}{1}\)
\(A=\frac{x-1}{\sqrt{x}}\)
Học tốt
2 tháng 8 2019
\(P=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(-\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)^2\)
\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(\frac{1}{4x}+\frac{1}{4}-\frac{1}{2}\right)\)
\(P=-\frac{4\sqrt{x}.\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=-\frac{4.\frac{x^2-2x+1}{4x}.\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(P=-\frac{\frac{x^2-2x+1}{\sqrt{x}}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=-\frac{x^2-2x+1}{\sqrt{x}.\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=-\frac{\sqrt{x}.\left(x-1\right)}{x}\)