Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2 xét x=0 => A =0
xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)
để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)
=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?
1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)
=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(a+2=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)
\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)
=> \(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)}+...\)
=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
=> M=0
Vậy M=0
a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)
c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)
b)
)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
= \(\frac{2}{2-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\)
=\(\frac{2\left(2+\sqrt{5}\right)-2\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)
=\(\frac{4+2\sqrt{5}-4+2\sqrt{5}}{2^2-\sqrt{5}^2}\)
=\(\frac{4\sqrt{5}}{4-5}\)
=\(\frac{4\sqrt{5}}{-1}\)
\(-4\sqrt{5}\)
a)Ta có: \(2\sqrt{5}< 5\sqrt{2}\)\(2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}\)
\(5\sqrt{2}=\sqrt{5^2.2}=\sqrt{50}\)
Vì \(\sqrt{20}< \sqrt{50}\)
Nên \(2\sqrt{5}< 5\sqrt{2}\)
b)Ta có: \(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(4\sqrt{11}=\sqrt{4^2.11}=\sqrt{176}\)
Vì \(\sqrt{117}< \sqrt{176}\)
Nên \(3\sqrt{13}< 4\sqrt{11}\)
c) Ta có: \(\frac{3}{4}.\sqrt{7}=\sqrt{\left(\frac{3}{4}\right)^2.7}=\sqrt{\frac{63}{16}}\)
\(\frac{2}{5}.\sqrt{5}=\sqrt{\left(\frac{2}{5}\right)^2.5}=\sqrt{\frac{4}{5}}\)
Vì \(\sqrt{\frac{63}{16}}>1\)
\(\sqrt{\frac{4}{5}}< 1\)
Nên \(\sqrt{\frac{63}{16}}>\sqrt{\frac{4}{5}}\)
Vậy \(\frac{3}{4}.\sqrt{7}>\frac{2}{5}.\sqrt{5}\)