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

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 $x=y=z=\dfrac{2}{3}$.

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

\(M=\frac{-4\left(x-8\sqrt{x}+16\right)+4x-28\sqrt{x}+49}{\left(\sqrt{x}-4\right)^2}=-4+\frac{\left(2\sqrt{x}-7\right)^2}{\left(\sqrt{x}-4\right)^2}\ge-4\)

Dấu "=" xảy ra khi \(2\sqrt{x}=7\Rightarrow x=\frac{49}{4}\)

Ta co:

\(M=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{x+\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-2}=\frac{x}{\sqrt{x}-2}=8+\frac{\left(\sqrt{x}-4\right)^2}{\sqrt{x}-2}\ge8\)

Dau '=' xay ra khi \(x=16\)

Vay \(M_{min}=8\)khi \(x=16\)