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9 tháng 8 2016
Nhầm 10√5 - 5√10 mới đúng
9 tháng 8 2016
10√5 + 5√10
8 tháng 10 2017

\(=\left(\frac{x-2\sqrt{x}-1}{x-4}-\frac{x-4}{x-4}\right):\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right]\)

\(=\frac{x-2\sqrt{x}-1-x+4}{x-4}:\left[\frac{\sqrt{x}-2}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right]\)

\(=\frac{3-2\sqrt{x}}{x-4}:\frac{\left(x-4\right)\left(\sqrt{x}-3\right)+\left(x-4\right)\left(\sqrt{x}+3\right)-\left(x-9\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)\left(\sqrt{x}+2\right)}\)

bạn làm tiếp nha! làm bằng máy tính phức tạp lắm

17 tháng 1 2020
https://i.imgur.com/ijO56jg.jpg

Tham khảo nhé!

Hỏi đáp Toán

11 tháng 12 2016

1/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)

\(\Leftrightarrow\frac{a+b+c}{abc}=0\)(đúng)

Vậy ta có ĐPCM

11 tháng 12 2016

2/ \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2006}-\sqrt{2005}\)

\(=\sqrt{2006}-1\)

18 tháng 8 2015

C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)

1 tháng 8 2019

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{a+1-a}=\sqrt{a+1}-\sqrt{a}\Rightarrow\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.......+\frac{1}{\sqrt{99}+\sqrt{100}}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-......-\sqrt{99}+\sqrt{100}=10-1=9\)

21 tháng 9 2017

\(\sqrt{19+8\sqrt{3}}-\sqrt{19-8\sqrt{3}}\)

\(=\sqrt{4^2+8\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{4^2-8\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{3}+4\right)^2}-\sqrt{\left(\sqrt{3}-4\right)^2}\)

\(=\left|\sqrt{3}+4\right|-\left|\sqrt{3}-4\right|\)

\(=\sqrt{3}+4-\sqrt{3}+4\)

\(=8\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

\(=\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}+1^2}+\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+1^2}\)

\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)

\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

11 tháng 11 2018

\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)

\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)

\(=x-2\sqrt{x}+4\)

=.= hok tốt!!