Tìm giá trị lớn nhất của \(\frac{2020-x}{6-x}\)

Ta có : \(\frac{2020-x}{6-x}=\frac{6-x+2014}{6-x}=\frac{6-x}{6-x}+\frac{2014}{6-x}=1+\frac{2014}{6-x}\)

Đa thức lớn nhất \(\Leftrightarrow1+\frac{2014}{6-x}\)lớn nhất  \(\Rightarrow\frac{2014}{6-x}\)lớn nhất  \(\Rightarrow6-x\)nhỏ nhất và \(6-x>0\)

Mà \(x\in Z\)\(\Rightarrow x=5\)

Vậy giá trị lớn nhất của đa thức \(=\frac{2020-5}{6-5}=2020-5=2015\)\(\Leftrightarrow x=5\)

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

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

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

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

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

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

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

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

(với 0<x<3) nha mn

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

\(% 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}\)

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

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

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

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

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

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

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

a,ĐKXĐ \(x\ne-1;-\frac{1}{2}\)

Ta thấy x=0 không là nghiệm của PT

Xét \(x\ne0\)

Khi đó PT 

<=> \(\frac{2}{6x-1+\frac{3}{x}}+\frac{5}{4x+5+\frac{2}{x}}+\frac{1}{2x+3+\frac{1}{x}}=\frac{1}{3}\)

Đặt \(2x+\frac{1}{x}=a\)

=> \(\frac{2}{3a-1}+\frac{5}{2a+5}+\frac{1}{a+3}=\frac{1}{3}\)

<=>  \(3\left(25a^2+75a+10\right)=6a^3+31a^2+34a-15\)

<=> \(6a^3-44a^2-191a-45=0\)

Xin lỗi đến đây tớ ra nghiệm không đẹp 

c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\)   ĐKXĐ \(x\ne-3\)

<=> \(\left(x-\frac{3x}{x+3}\right)^2+2.\frac{3x^2}{x+3}=7\)

<=> \(\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-7=0\)

<=> \(\left(\frac{x^2}{x+3}+7\right)\left(\frac{x^2}{x+3}-1\right)=0\)

<=> \(\orbr{\begin{cases}x^2+7x+21=0\\x^2-x-3=0\end{cases}}\)

\(S=\left\{\frac{1\pm\sqrt{13}}{2}\right\}\)thỏa mãn ĐKXĐ