HN
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
P
2
11 tháng 9 2016
\(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{23}\)
29 tháng 10 2018
Đặt \(A=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{\left(a+\sqrt{a^2-b^2}\right)\left(a-\sqrt{a^2-b^2}\right)}{b\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(A=\frac{a-b}{\sqrt{a-b}.\sqrt{a+b}}\)
\(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
Với \(a=3b\) ta có : \(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}}{2}\)
Chúc bạn học tốt ~
RA
1
7 tháng 10 2020
Ta có: \(\sqrt{18}-\frac{1}{3}\sqrt{72}-\sqrt{8}+\frac{2-3\sqrt{2}}{3-\sqrt{2}}\)
\(=3\sqrt{2}-\frac{6\sqrt{2}}{3}-2\sqrt{2}+\frac{\left(3+\sqrt{2}\right)\left(2-3\sqrt{2}\right)}{9-2}\)
\(=3\sqrt{2}-2\sqrt{2}-2\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
11 tháng 3 2022
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)
\(A=\dfrac{1}{2\left(\sqrt{a}+1\right)}-\dfrac{1}{2\left(\sqrt{a}-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{\sqrt{a}-1-\sqrt{a}-1}{2\left(a-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-1}{a-1}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-a-1+a^2+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{a^2-a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a+1}\)
b: Để A-1/3<0 thì \(\dfrac{a}{a+1}-\dfrac{1}{3}< 0\)
=>3a-a-1<0
=>2a-1<0
hay 0<a<1/2
14 tháng 2 2019
\(x=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\)
\(\Leftrightarrow x+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(\Leftrightarrow\left(x+\frac{\sqrt{2}}{8}\right)^2=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\)
\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)
\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}-\frac{\sqrt{2}}{4}=0\)
\(\Leftrightarrow4x^2+x\sqrt{2}-\sqrt{2}=0\)(1)
\(\Leftrightarrow x\sqrt{2}=\sqrt{2}-4x^2\)
\(\Leftrightarrow x=1-2x^2\sqrt{2}\)
Thay vào M ta sẽ được
\(M=x^2+\sqrt{x^4+1-2x^2\sqrt{2}+1}\)
\(=x^2+\sqrt{\left(x^2-\sqrt{2}\right)^2}\)
\(=x^2+\left|x^2-\sqrt{2}\right|\)
Từ \(\left(1\right)\Rightarrow\sqrt{2}-x\sqrt{2}=4x^2\ge0\)
\(\Leftrightarrow\sqrt{2}\left(1-x\right)\ge0\)
\(\Leftrightarrow x\le1\)
\(\Leftrightarrow x^2\le1< \sqrt{2}\)
\(\Rightarrow\left|x^2-\sqrt{2}\right|=\sqrt{2}-x^2\)
Khi đó \(M=x^2+\left|x^2-\sqrt{2}\right|=x^2-\sqrt{2}+x^2=\sqrt{2}\)
|N|
SN
1
VT
15 tháng 9 2017
ta có ĐK là x>=0
ta có \(4\sqrt{x}\ge0;x+2\sqrt{x}+1>0\Rightarrow\) \(\frac{4\sqrt{x}}{x+2\sqrt{x}+1}\ge0\)
dấu = xảy ra <=> x= 0,
AD
1
T
15 tháng 9 2018
\(=\frac{\left(\sqrt{x}-\sqrt{4y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}+\frac{3x.\left(x-\sqrt{xy}\right)}{\left(x+\sqrt{xy}\right).\left(x-\sqrt{xy}\right)}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3x.\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x^2-xy}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3x\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x.\left(x-y\right)}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)+3.\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}-2.\sqrt{y}+3.\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=1\)