- rut gon
a, \(\frac{132639}{173451}\) b, \(\frac{16515}{20919}\)
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20 tháng 2 2017
\(A=\frac{132639}{173451}\)
\(A=\frac{132639:10203}{173451:10203}\)
\(A=\frac{13}{17}\)
\(B=\frac{16515}{20919}\)
\(B=\frac{16515:1101}{20919:1101}\)
\(B=\frac{15}{19}\)
HG
2
5 tháng 10 2017
A=\(\frac{a\sqrt{a}+b\sqrt{b}-\left(a+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}\)=\(\frac{-a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}\)=\(\frac{-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}\)=\(-1\)
NQ
0
PR
1
30 tháng 12 2019
\(\frac{a-b}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a}^3+\sqrt{b}^3}{a-b}\)
\(=\sqrt{a}+\sqrt{b}+\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}+\frac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}+\frac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{a-b+a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{2a-\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
1 tháng 8 2019
\(\sqrt{\frac{a}{b}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{b}{a}}\)
\(=\sqrt{\frac{a}{b}}+\sqrt{ab}+\sqrt{\frac{a^2b}{b^2a}}\)
\(=\sqrt{\frac{a}{b}}+\sqrt{ab}+\sqrt{\frac{a}{b}}\)
\(=2\sqrt{\frac{a}{b}}+\sqrt{ab}\)
11 tháng 10 2017
tham khao nha
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right):\left(\frac{\sqrt{b}+\sqrt{a}}{\sqrt{ab}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{a-2\sqrt{ab}+b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
vay \(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
11 tháng 10 2017
ĐK : tự ghi nha
\(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
ta co :a)132639/173451=132639:10203/173451:10203=13/17
b)16515/20919=16515:1101/20919:1101=15/19
13/17
15/19