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10 tháng 8 2017
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
10 tháng 8 2017
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
HN
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 10 2020
ĐKXĐ: ...
a.
\(VT=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{2\left(6-x+x+2\right)}=4\)
\(VP=\left(x-3\right)^2+4\ge4\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}6-x=x+2\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) (ko tồn tại x thỏa mãn)
Vậy pt vô nghiệm
b.
\(\Leftrightarrow\frac{x^3}{\sqrt{9-x^2}}=9-x^2\)
\(\Leftrightarrow x^3=\left(9-x^2\right)\sqrt{9-x^2}\)
\(\Leftrightarrow x^3=\left(\sqrt{9-x^2}\right)^3\)
\(\Leftrightarrow x=\sqrt{9-x^2}\) (\(x\ge0\))
\(\Leftrightarrow x^2=9-x^2\)
\(\Rightarrow x=\frac{3\sqrt{2}}{2}\)
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}\)
T
17 tháng 7 2019
\(6-2x-\sqrt{9-6x+x^2}\)
= \(6-2x-\sqrt{\left(3-x\right)^2}\)
= \(\left\{{}\begin{matrix}6-2x-3+x\\6-2x+3-x\end{matrix}\right.\)
= \(\left\{{}\begin{matrix}3-x\\9-3x\end{matrix}\right.\)
\(\frac{3-\sqrt{x}}{x-9}\)
=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(x-3\right)}\)
= \(\frac{-1}{\sqrt{x}+3}\)
HN
19 tháng 8 2016
a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
Đặt \(t=\sqrt{x-1}\left(ĐK:t\ge0\right)\Leftrightarrow x-1=t^2\Leftrightarrow x=t^2+1\)
pt \(\Leftrightarrow\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t}=2\Leftrightarrow\sqrt{\left(t+1\right)^2}+\sqrt{\left(t-1\right)^2}=2\Leftrightarrow t+1+t-1=2\Leftrightarrow t=1\left(tm\right)\)
Với t=1 \(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)
Câu b tương tự
28 tháng 7 2017
e, \(E=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(E\ge\left|x-1+3-x\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Rightarrow1\le x\le3\)
Vậy \(MIN_E=2\) khi \(1\le x\le3\)
f, \(F=\sqrt{x+9-6\sqrt{x}}+\sqrt{x+1-2\sqrt{x}}\)
\(=\sqrt{\left(\sqrt{x}-3\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\)
\(=\left|\sqrt{x}-3\right|+\left|\sqrt{x}-1\right|=\left|3-\sqrt{x}\right|+\left|\sqrt{x}-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(F\ge\left|3-\sqrt{x}+\sqrt{x}-1\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}3-\sqrt{x}\ge0\\\sqrt{x}-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{3}\\x\ge1\end{matrix}\right.\)
Vậy \(MIN_F=2\) khi \(1\le x\le\sqrt{3}\)