\(\frac{\sqrt{4a}}{x-3}\) . \(\sqrt{\frac{x^2-6x+9}{x}...">
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10 tháng 10 2020

(với 0<x<3) nha mn

10 tháng 10 2020

\(=\frac{2\sqrt{x}}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{x}}=\frac{2\left(x-3\right)}{x-3}=-2\)

21 tháng 9 2017

\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{x^2-6x+9}}{\sqrt{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{\left(2y^2\right)^2}}\)

\(=\frac{2}{x-3}.\frac{x-3}{2y^2}=\frac{1}{y^2}\)

21 tháng 9 2017

\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}\)

\(=\frac{2}{x-3}\sqrt{\frac{\left(x-3\right)^2}{\left(2y^2\right)^2}}\)

\(=\frac{2}{x-3}.\left|\frac{x-3}{2y^2}\right|\)

\(=\frac{2}{x-3}.\frac{3-x}{2y^2}\)( vi \(x< 3;y\ne0\))

\(=\frac{-2}{2y^2}\)

\(=\frac{-1}{y^2}\)

17 tháng 6 2019

\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)

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

                                                                               \(=y-x\)

\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)

\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)

17 tháng 6 2019

\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)

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

\(=|\sqrt{x}^2-\sqrt{y}^2|\)

\(=|x-y|\)

Vì \(x\le y\)\(\Rightarrow x-y\ge0\)

\(\Rightarrow|x-y|=x-y\)

28 tháng 7 2016

a)\(x+3+\sqrt{x^2-6x+9}\)

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

\(=x+3+x-3\)

\(=2x\)

b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)

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

\(=x+2-x\)

=2

c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)

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

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

14 tháng 7 2016

a/ \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

     \(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

       \(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

b/ \(P< 1\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}< 1\Rightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

   Xét 2 trường hợp:

  • \(\hept{\begin{cases}2\sqrt{x}+3>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\hept{\begin{cases}2\sqrt{x}>-3\\\sqrt{x}< 3\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{x}>-\frac{3}{2}\\\sqrt{x}< 3\end{cases}}\Rightarrow-\frac{3}{2}< \sqrt{x}< 3}\)

                                       \(\Rightarrow-\frac{9}{4}< x< 9\)

  •  \(\hept{\begin{cases}2\sqrt{x}+3< 0\\\sqrt{x}>3\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x}< -\frac{3}{2}\\\sqrt{x}>3\end{cases}}}\) (vô lí)

                                                   Vậy -9/4 < x < 9

17 tháng 7 2019

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaaI2a % GaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaaiaaiMdacqGHsisl % caaI2aGaamiEaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaabe % aakmaabmaabaGaamiEaiabgYda8iaaiodaaiaawIcacaGLPaaaaeaa % cqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaam % aabmaabaGaaG4maiabgkHiTiaadIhaaiaawIcacaGLPaaadaahaaWc % beqaaiaaikdaaaaabeaaaOqaaiabg2da9iaaiAdacqGHsislcaaIYa % GaamiEaiabgkHiTmaaemaabaGaaG4maiabgkHiTiaadIhaaiaawEa7 % caGLiWoaaeaacqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHRa % WkcaaIZaGaeyOeI0IaamiEaaqaaiabg2da9iaaiMdacqGHsislcaaI % ZaGaamiEaaqaamaalaaabaGaaG4maiabgkHiTmaakaaabaGaamiEaa % WcbeaaaOqaaiaadIhacqGHsislcaaI5aaaamaabmaabaGaamiEaiab % gwMiZkaaicdacaGGSaGaamiEaiabgcMi5kaaiMdaaiaawIcacaGLPa % aaaeaacqGH9aqpdaWcaaqaaiabgkHiTmaabmaabaWaaOaaaeaacaWG % 4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMcaaaqaamaabmaaba % WaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMca % amaabmaabaWaaOaaaeaacaWG4baaleqaaOGaey4kaSIaaG4maaGaay % jkaiaawMcaaaaaaeaacqGH9aqpdaWcaaqaaiabgkHiTiaaigdaaeaa % daGcaaqaaiaadIhaaSqabaGccqGHRaWkcaaIZaaaaaqaamaalaaaba % GaamiEaiabgkHiTiaaiwdadaGcaaqaaiaadIhaaSqabaGccqGHRaWk % caaI2aaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaaada % qadaqaaiaadIhacqGHLjYScaaIWaGaaiilaiaadIhacqGHGjsUcaaI % 5aaacaGLOaGaayzkaaaabaGaeyypa0ZaaSaaaeaacaWG4bGaeyOeI0 % IaaGOmamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaiodadaGcaaqa % aiaadIhaaSqabaGccqGHRaWkcaaI2aaabaWaaOaaaeaacaWG4baale % qaaOGaeyOeI0IaaG4maaaaaeaacqGH9aqpdaWcaaqaamaakaaabaGa % amiEaaWcbeaakmaabmaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaaiabgkHiTiaaiodadaqadaqaamaakaaa % baGaamiEaaWcbeaakiabgkHiTiaaikdaaiaawIcacaGLPaaaaeaada % GcaaqaaiaadIhaaSqabaGccqGHsislcaaIZaaaaaqaaiabg2da9maa % laaabaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGHsislcaaIYa % aacaGLOaGaayzkaaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGH % sislcaaIZaaacaGLOaGaayzkaaaabaWaaOaaaeaacaWG4baaleqaaO % GaeyOeI0IaaG4maaaaaeaacqGH9aqpdaGcaaqaaiaadIhaaSqabaGc % cqGHsislcaaIYaaaaaa!C78C! \begin{array}{l} 6 - 2x - \sqrt {9 - 6x + {x^2}} \left( {x < 3} \right)\\ = 6 - 2x - \sqrt {{{\left( {3 - x} \right)}^2}} \\ = 6 - 2x - \left| {3 - x} \right|\\ = 6 - 2x + 3 - x\\ = 9 - 3x\\ \dfrac{{3 - \sqrt x }}{{x - 9}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{ - \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{ - 1}}{{\sqrt x + 3}}\\ \dfrac{{x - 5\sqrt x + 6}}{{\sqrt x - 3}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{x - 2\sqrt x - 3\sqrt x + 6}}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - 3\left( {\sqrt x - 2} \right)}}{{\sqrt x - 3}}\\ = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\ = \sqrt x - 2 \end{array}\)

17 tháng 7 2019

\(6-2x-\sqrt{9-6x+x^2}\)

= \(6-2x-\sqrt{\left(3-x\right)^2}\)

= \(\left\{{}\begin{matrix}6-2x-3+x\\6-2x+3-x\end{matrix}\right.\)

= \(\left\{{}\begin{matrix}3-x\\9-3x\end{matrix}\right.\)

\(\frac{3-\sqrt{x}}{x-9}\)

=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(x-3\right)}\)

= \(\frac{-1}{\sqrt{x}+3}\)