\(P=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(P=\left(x^2+5x\right)^2-36\)

\(P=\left[x\left(x+5\right)\right]^2-36\)

Vậy GTNN của P = -36 khi x = 0 hoặc -5.

1. A=x²+6x+10=x^2+2.3x+9+1=(x+3)²+1 dat gia tri nho nhat la 1 khi do x=-3

\(y=\left|x^2+x+16\right|+\left|x^2+x-6\right|=\left|x^2+x+16\right|+\left|6-x-x^2\right|\)

\(\ge\left|x^2+x+16+6-x-x^2\right|=22\)

Dấu m"=" xảy ra <=> \(-16\le x^2+x\le6\)

<=> \(-3\le x\le2\)

Vậy giá trị nhỏ nhất của y = 22 đạt tại  \(-3\le x\le2\)

\(\left\{{}\begin{matrix}4x^2+9y^2=9\\A=x-2y+3\end{matrix}\right.\)

Áp dụng bất đẳng thức Bunhiacopxki cho các cặp số \(\left(\dfrac{1}{2};2x\right);\left(-\dfrac{2}{3};3y\right)\)

\(x-2y=\dfrac{1}{2}.x+\left(-\dfrac{2}{3}\right).3y\)

\(\Rightarrow\left[\dfrac{1}{2}.2x+\left(-\dfrac{2}{3}\right).3y\right]^2\le\left(\dfrac{1}{4}+\dfrac{4}{9}\right)\left(4x^2+9y^2\right)=\dfrac{25}{36}.9\)

\(\Rightarrow x-2y\le\dfrac{5}{6}.3=\dfrac{5}{2}\)

\(\Rightarrow A=x-2y+3\le\dfrac{5}{2}+3\)

\(\Rightarrow A=x-2y+3\le\dfrac{11}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{\dfrac{1}{2}}{2x}=\dfrac{-\dfrac{2}{3}}{3y}\)

\(\Rightarrow\dfrac{2x}{\dfrac{1}{2}}=\dfrac{3y}{-\dfrac{2}{3}}\)

\(\Rightarrow\dfrac{4x^2}{\dfrac{1}{4}}=\dfrac{9y^2}{\dfrac{4}{9}}=\dfrac{4x^2+9y^2}{\dfrac{1}{4}+\dfrac{4}{9}}=\dfrac{9}{\dfrac{25}{36}}=\dfrac{9.36}{25}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{9.36}{25}.\dfrac{1}{16}\\y^2=\dfrac{9.36}{25}.\dfrac{4}{36}=\dfrac{9.4}{25}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.6}{5}.\dfrac{1}{4}=\dfrac{9}{10}\\y=\dfrac{3.2}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy \(GTLN\left(A\right)=\dfrac{11}{2}\left(tạix=\dfrac{9}{10};y=\dfrac{6}{5}\right)\)

\(D=\frac{x^2-2x+2014}{x^2}\)

\(D=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2014}{x^2}\)

\(D=1-\frac{2}{x}+\frac{2014}{x^2}\)

\(D=2014\cdot\frac{1}{x^2}-2\cdot\frac{1}{x}+1\)

Đặt \(\frac{1}{x}=a\)

\(D=2014a^2-2a+1\)

\(D=2014\left(a^2-a\cdot\frac{1}{1007}+\frac{1}{2014}\right)\)

\(D=2014\left(a^2-2\cdot a\cdot\frac{1}{2014}+\frac{1}{2014^2}+\frac{2013}{2014^2}\right)\)

\(D=2014\left[\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)

\(D=2014\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2014}\Leftrightarrow\frac{1}{x}=\frac{1}{2014}\Leftrightarrow x=2014\)

Vậy....

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

\(=1-\frac{1}{x+1}+\frac{1}{\left(x+1\right)^2}=\left[\frac{1}{4}-\frac{1}{x+1}+\frac{1}{\left(x+1\right)^2}\right]+\frac{3}{4}\)

\(=\left(\frac{1}{2}-\frac{1}{x+1}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow P\ge\frac{3}{4}\)

Vậy \(Max_P=\frac{3}{4}\Leftrightarrow x=1\)

Ta có :

\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\forall x\)

\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu bằng xảy ra khi và chỉ khi :
\(\left(x^2+5x\right)^2=0\)

\(\Leftrightarrow x^2+5x=0\)

\(x\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy \(P_{min}=-36\)tại \(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)