Câu 1Rút...
Đọc tiếp
Câu 1
Rút gọn:
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
F
20 tháng 8 2021
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
11 tháng 10 2021
a: Ta có: \(A=\left(1-\dfrac{2\sqrt{x}-2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{x-1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)
13 tháng 7 2021
Ta có: S=A-B
\(=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
18 tháng 9 2021
1)\(\sqrt{3-2\sqrt{2}}-\sqrt{2}=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2}=\sqrt{2}-1-\sqrt{2}=-1\left(đpcm\right)\)
2) \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
3) \(ĐK:\)\(\left\{{}\begin{matrix}\dfrac{x-1}{x+3}\ge0\\x+3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\\x\ne-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x< -3\end{matrix}\right.\)
4) \(ĐK:\left\{{}\begin{matrix}7-x\ge0\\a\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\a\ge0\end{matrix}\right.\)
11 tháng 7 2021
Ta có: M=A+B
\(=\dfrac{x-\sqrt[3]{x}}{x-1}+\dfrac{1}{\sqrt[3]{x}-1}+\dfrac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
\(=\dfrac{x-\sqrt[3]{x}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}+\dfrac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\dfrac{x+\sqrt[3]{x}+\sqrt[3]{x^2}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)
8 tháng 7 2021
\(M=\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) với x>0;x≠1
\(=\left(\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\)
\(M=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{x}=\dfrac{-x+\sqrt{x}+2}{x\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{2-\sqrt{x}}{x}\)
vậy M=\(\dfrac{2-\sqrt{x}}{x}\)
vì x>0 nên để \(M< 0\Leftrightarrow\dfrac{2-\sqrt{x}}{x}< 0\Leftrightarrow2-\sqrt{x}< 0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
20 tháng 9 2021
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
rút gọn gì?
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}=15\sqrt{5a}+\sqrt{a}\)