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1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
x^2+x+1/4+3/4
=(x+1/2)^2+3/4
=> A min=3/4
Câu kia tương tự .......
\(A=x^2+x+1=x^2+2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0,x\in R\)
nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},x\in R\)
Vậy \(Min_A=\frac{3}{4}\)khi \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
\(B=\left(x+2\right)^2+\left(x-3\right)^2=x^2+2x+1+x^2-6x+9=2x^2-4x+10=2\left(x^2-2x+5\right)\)
\(B=2\left(x^2-2x+1+4\right)=2\left(x-1\right)^2+4\)
Vì \(2\left(x-1\right)^2\ge0,x\in R\)
nên \(2\left(x-1\right)^2+4\ge4,x\in R\)
Vậy \(Min_B=4\)khi \(x-1=0\Rightarrow x=1\)
\(P=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)\)
\(=\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)\)
\(=\left(x^2+3x+1\right)^2-1\ge-1\)
Vậy GTNN của P là -1 khi \(\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{5}}{2}\\x_2=\dfrac{-3-\sqrt{5}}{2}\end{matrix}\right.\)
\(P=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
\(P=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]\)
\(P=\left(x^2+3x\right)\left(x^2+3x+2\right)\)
Đặt x2 + 3x + 1 = a, ta được:
\(P=\left(a-1\right)\left(a+1\right)\)
\(P=a^2-1\)
Vì \(a^2\ge0\) với mọi a
\(\Rightarrow a^2-1\ge-1\)
\(\Rightarrow Pmin=-1\)\(\Leftrightarrow a=0\)
\(\Leftrightarrow x^2+3x+1=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}+1=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\sqrt{\dfrac{5}{4}}\\x+\dfrac{3}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{3}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}}{2}-\dfrac{3}{2}\\x=-\dfrac{\sqrt{5}}{2}-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{5}}{2}\\x=-\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\)