B = \(\dfrac{x^2+15}{x^2+3}\)

= \(\dfrac{x^2+3+12}{x^2+3}\)

= 1 + \(\dfrac{12}{x^2+3}\)

Để B đạt GTLN

=> x² + 3 nhỏ nhất

Mà x² nhỏ nhất bằng 0

=> x² + 3 nhỏ nhất bằng 3

Vậy GTLN của B là 1 + \(\dfrac{12}{3}\)

= 1 + 4

= 5

x^2+6x+12=(x+3)^2+3>=3

=>B<=5/3

Dấu = xảy ra khi x=-3

B=(x+1)(x+4)(x+2)(x+3)\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

đặt x^2+5x+5 =t (t>=0)

=> B=\(\left(t+1\right)\left(t-1\right)=t^2-1\) ta có: \(t^2\ge0\Rightarrow t^2-1\ge-1\Rightarrow MinB=-1\Leftrightarrow t=0\Leftrightarrow x^2+5x+5=0\Leftrightarrow\left(x^2+5x+\frac{25}{4}\right)=\frac{5}{4}\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\Rightarrow x=-\frac{5}{2}+-\frac{\sqrt{5}}{2}\)

B=(x+1)(x+2)(x+3)(x+4)T=(x+1)(x+2)(x+3)(x+4)

=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+4)(x+2)(x+3)

=(x2+5x+4)(x2+5x+6)=(x2+5x+4)(x2+5x+6)

Đặt :x^2+5x+4=ax2+5x+4=a ⇒T=(a−1)(a+1)⇒T=(a−1)(a+1)

=a^2−1=(x2+5x+5)2−1≥−1=a2−1=(x2+5x+5)2−1≥−1

Vậy MinT=−1MinT=−1 khi

x2+5x+5=0⇒(x2+5x+254)−54=0x2+5x+5=0⇒(x2+5x+254)−54=0⇔(x+52)2=54⇔(x+52)2=54

\(\Rightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}-\frac{5}{2}}\\x=-\sqrt{\frac{5}{4}-\frac{5}{2}}\end{cases}}\)

B=(x+1)(x+2)(x+3)(x+4)T=(x+1)(x+2)(x+3)(x+4)

=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+4)(x+2)(x+3)

=(x2+5x+4)(x2+5x+6)=(x2+5x+4)(x2+5x+6)

Đặt x2+5x+4=ax2+5x+4=a ⇒B=(a−1)(a+1)⇒T=(a−1)(a+1)

=a2−1=(x2+5x+5)2−1≥−1=a2−1=(x2+5x+5)2−1≥−1

Vậy MinB=−1MinB=−1 khi

x2+5x+5=0⇒(x2+5x+254)−54=0x2+5x+5=0⇒(x2+5x+254)−54=0⇔(x+52)2=54⇔(x+52)2=54

⇒⎡⎣⎢⎢⎢x+52=54−−√x+52=−54−−√⇒⎡⎣⎢⎢⎢x=54−−√−52x=−54−−√−52

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