Rút gọn biểu thức:
B = \(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\) ( ĐK: \(x\ge0\); \(x\ne1\) )
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WJ
9 tháng 8 2016
Q=\(\frac{\sqrt{x}-1}{x-\sqrt{x}+1}+\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\) điều kiện x>=0
=\(\frac{x-1+x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
ta thấy cả tử và mẫu đề >=0=> Q>=0
dấu = xảy ra khi x=0
=> Q=0 khi x=0
N
26 tháng 3 2019
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
2 tháng 11 2017
\(M=\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(M=\left(\frac{2x}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right):\frac{x+1+\sqrt{x}}{x+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\frac{x+\sqrt{x}+1}{x+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\frac{x+1}{x+\sqrt{x}+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(M=\frac{2x}{x^3-1}\)
vậy \(M=\frac{2x}{x^3-1}\)
9 tháng 3 2020
\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
2, Với x>1 ta có \(\frac{1}{A}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\frac{3}{\sqrt{x}-1}+3\)
Áp dụng bđt AM-GM ta có
\(\frac{1}{A}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{3}{\sqrt{x}-1}}+3=2\sqrt{3}+3\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
6 tháng 7 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
6 tháng 7 2019
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
BQ
17 tháng 9 2018
Chào em, em có thể kam khảo tại link:
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
Nếu link bị chặn em copy và dán tại:
https://olm.vn/hoi-dap/question/1261852.html
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
D
17 tháng 9 2018
a) Rút gọn E
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-\sqrt{x}}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}-\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
Vậy \(E=\frac{x}{\sqrt{x}-1}\)
\(B=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(x-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x^2+x\sqrt{x}+\sqrt{x}+1-x^2+2x-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x\sqrt{x}+\sqrt{x}+2x}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(x+1+2\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
tiếp tục của bạn @Bastkoo nhé
\(B=\frac{\sqrt{x}\left(x+2\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}}{\sqrt{x}-1}\)