\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

=\(\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

Cám ơn bạn mã mã , để mình làm nốt nhé :

\(A=\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

Để \(A>2\Rightarrow\frac{4\sqrt{x}+8}{\sqrt{x}+7}>2\)

\(\Rightarrow\frac{4\sqrt{x}+8}{\sqrt{x}+7}-2>0\)

\(\Rightarrow\frac{4\sqrt{x}+8-2\sqrt{x}-14}{\sqrt{x}+7}>0\)

\(\Rightarrow\frac{2\sqrt{x}-6}{\sqrt{x}+7}>0\)

Vì \(\sqrt{x}>0\Rightarrow\sqrt{x}+7>0\)\(\Rightarrow A>0\Leftrightarrow2\sqrt{x}-6>0\)

\(\Rightarrow2\left(\sqrt{x}-3\right)>0\Rightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow\sqrt{x}>3\Rightarrow\sqrt{x}>\sqrt{9}\Rightarrow x>9\)

Vậy để \(A>2\Leftrightarrow x>9\)

\(1)\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} - \sqrt 6 } \right)\left( {\sqrt {4 - \sqrt {15} } } \right)\\ = \left( {4\sqrt {10} - 4\sqrt 6 + \sqrt {150} - \sqrt {90} } \right)\sqrt {4 - \sqrt {15} } \\ = \left( {4\sqrt {10} - 4\sqrt 6 + 5\sqrt 6 - 3\sqrt {10} } \right)\sqrt {4 - \sqrt {15} } \\ = \left( {\sqrt {10} + \sqrt 6 } \right)\sqrt {4 - \sqrt {15} } \\ = \sqrt {10\left( {4 - \sqrt {15} } \right)} + \sqrt {6\left( {4 - \sqrt {15} } \right)} \\ = \sqrt {40 - 10\sqrt {15} } + \sqrt {24 - 6\sqrt {15} } \\ = \sqrt {{{\left( {5 - \sqrt {15} } \right)}^2}} + \sqrt {{{\left( {3 - \sqrt {15} } \right)}^2}} \\ = 5 - \sqrt {15} + \sqrt {15} - 3 = 2\)

2) Áp dụng bất đẳng thức AM - GM ta có

\(\dfrac{{{x^2}}}{{y + z}} + \dfrac{{y + z}}{4} \ge 2\sqrt {\dfrac{{{x^2}}}{{y + z}}.\dfrac{{y + z}}{4}} = x(1)\)

Hoàn toàn tương tự:

\(\dfrac{{{y^2}}}{{z + x}} + \dfrac{{z + x}}{4} \ge y\left( 2 \right)\\ \dfrac{{{z^2}}}{{x + y}} + \dfrac{{x + y}}{4} \ge z\left( 3 \right) \)

Từ (1), (2), (3) ta có ngay:\(\left(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\right)+ \left(\dfrac{y^2}{z+x}+\dfrac{z+x}{4}\right)+\left( \dfrac{z^2}{x+y} +\dfrac{x+y}{4}\right)\geqslant x+y+z\\ \iff\dfrac{x^2}{y+z}+ \dfrac{y^2}{z+x}+ \dfrac{z^2}{x+y}\geqslant \dfrac{x+y+z}{2} \)

Chú ý rằng \(x+y+z=2\), ta có ngay\(\dfrac{x^2}{y+z}+ \dfrac{y^2}{z+x}+ \dfrac{z^2}{x+y}\geqslant 1\)

Vậy giá trị nhỏ nhất của $P$ là $1$, đạt được khi $x=y=z=\dfrac{2}{3}$.

Haizzz bị lỗi công thức suốt :((

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaaIXa % GaaiykamaabmaabaGaaGinaiabgUcaRmaakaaabaGaaGymaiaaiwda % aSqabaaakiaawIcacaGLPaaadaqadaqaamaakaaabaGaaGymaiaaic % daaSqabaGccqGHsisldaGcaaqaaiaaiAdaaSqabaaakiaawIcacaGL % PaaadaqadaqaamaakaaabaGaaGinaiabgkHiTmaakaaabaGaaGymai % aaiwdaaSqabaaabeaaaOGaayjkaiaawMcaaaqaaiabg2da9maabmaa % baGaaGinamaakaaabaGaaGymaiaaicdaaSqabaGccqGHsislcaaI0a % WaaOaaaeaacaaI2aaaleqaaOGaey4kaSYaaOaaaeaacaaIXaGaaGyn % aiaaicdaaSqabaGccqGHsisldaGcaaqaaiaaiMdacaaIWaaaleqaaa % GccaGLOaGaayzkaaWaaOaaaeaacaaI0aGaeyOeI0YaaOaaaeaacaaI % XaGaaGynaaWcbeaaaeqaaaGcbaGaeyypa0ZaaeWaaeaacaaI0aWaaO % aaaeaacaaIXaGaaGimaaWcbeaakiabgkHiTiaaisdadaGcaaqaaiaa % iAdaaSqabaGccqGHRaWkcaaI1aWaaOaaaeaacaaI2aaaleqaaOGaey % OeI0IaaG4mamaakaaabaGaaGymaiaaicdaaSqabaaakiaawIcacaGL % PaaadaGcaaqaaiaaisdacqGHsisldaGcaaqaaiaaigdacaaI1aaale % qaaaqabaaakeaacqGH9aqpdaqadaqaamaakaaabaGaaGymaiaaicda % aSqabaGccqGHRaWkdaGcaaqaaiaaiAdaaSqabaaakiaawIcacaGLPa % aadaGcaaqaaiaaisdacqGHsisldaGcaaqaaiaaigdacaaI1aaaleqa % aaqabaaakeaacqGH9aqpdaGcaaqaaiaaigdacaaIWaWaaeWaaeaaca % aI0aGaeyOeI0YaaOaaaeaacaaIXaGaaGynaaWcbeaaaOGaayjkaiaa % wMcaaaWcbeaakiabgUcaRmaakaaabaGaaGOnamaabmaabaGaaGinai % abgkHiTmaakaaabaGaaGymaiaaiwdaaSqabaaakiaawIcacaGLPaaa 1)\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} - \sqrt 6 } \right)\left( {\sqrt {4 - \sqrt {15} } } \right)\\ = \left( {4\sqrt {10} - 4\sqrt 6 + \sqrt {150} - \sqrt {90} } \right)\sqrt {4 - \sqrt {15} } \\ = \left( {4\sqrt {10} - 4\sqrt 6 + 5\sqrt 6 - 3\sqrt {10} } \right)\sqrt {4 - \sqrt {15} } \\ = \left( {\sqrt {10} + \sqrt 6 } \right)\sqrt {4 - \sqrt {15} } \\ = \sqrt {10\left( {4 - \sqrt {15} } \right)} + \sqrt {6\left( {4 - \sqrt {15} } \right)} \\ = \sqrt {40 - 10\sqrt {15} } + \sqrt {24 - 6\sqrt {15} } \\ = \sqrt {{{\left( {5 - \sqrt {15} } \right)}^2}} + \sqrt {{{\left( {3 - \sqrt {15} } \right)}^2}} \\ = 5 - \sqrt {15} + \sqrt {15} - 3 = 2 \)

2) Áp dụng bất đẳng thức AM - GM ta có

\begin{equation} \label{eq:1} \dfrac{x^2}{y+z}+\dfrac{y+z}{4}\geqslant 2\sqrt{\dfrac{x^2}{y+z}\cdot \dfrac{y+z}{4}}=x \end{equation}

Hoàn toàn tương tự:

\begin{align} \label{eq:2} \dfrac{y^2}{z+x}+\dfrac{z+x}{4}\geqslant y \\ \label{eq:3} \dfrac{z^2}{x+y}+\dfrac{x+y}{4}\geqslant z \end{align}

Từ \eqref{eq:1}, \eqref{eq:2}, \eqref{eq:3} ta có ngay

\[\left(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\right)+ \left(\dfrac{y^2}{z+x}+\dfrac{z+x}{4}\right)+\left( \dfrac{z^2}{x+y} +\dfrac{x+y}{4}\right)\geqslant x+y+z\]

\[\iff\dfrac{x^2}{y+z}+ \dfrac{y^2}{z+x}+ \dfrac{z^2}{x+y}\geqslant \dfrac{x+y+z}{2}\]

Chú ý rằng $x+y+z=2$, ta có ngay

\[\dfrac{x^2}{y+z}+ \dfrac{y^2}{z+x}+ \dfrac{z^2}{x+y}\geqslant 1\]

Vậy giá trị nhỏ nhất của $P$ là $1$, đạt được khi $x=y=z=\dfrac{2}{3}$.

Điều kiện có 2 nghiệm phân biệt tự làm nha

Theo vi-et ta có:

\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)

\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)

\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)

\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)

Làm nốt nhé

Câu 1:

M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)

=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)

=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)

\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)

\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)

2/ \(S=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=9\)