khó hiểu quá 

bn giải giúp mình đi

1)   P = \(3+15x-5x^2\)\(=-5x^2+15x+3=-5\left(x^2-3x-\frac{3}{5}\right)\)  \(=-5\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}-\frac{9}{4}-\frac{3}{5}\right)\)= \(-5\left[\left(x-\frac{3}{2}\right)^2-\frac{57}{20}\right]=-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}\)

vì \(\left(x-\frac{3}{2}\right)^2>=0\) => \(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=0\)  =>\(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=\frac{57}{4}\)

 => GTLN  của P là \(\frac{57}{4}\)tại x =\(\frac{3}{2}\)

2) GTNN của B là -36

gấp ạ, 10p nữa mình phải xong

\(A=\dfrac{x^2+3}{x^2+1}=1+\dfrac{2}{x^2+1}\le1+\dfrac{2}{1}=3\) 

" = " \(\Leftrightarrow x=0\)

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

Để A đạt GTLN thì \(x^2+1\) nhỏ nhất

Mà \(x^2+1\ge1;\forall x\) 

\(\Rightarrow A_{Max}=1+\dfrac{2}{1}=3\)

Dấu "=" xảy ra khi \(x^2=0\Leftrightarrow x=0\)

\(ĐKXĐ:x\ne-1\)

\(B=\frac{3\left(x+1\right)}{x^3+x^2+x+1}\)\(\Leftrightarrow B=\frac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}\)\(\Leftrightarrow B=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)\(\Leftrightarrow B=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\)\(\Leftrightarrow B=\frac{3}{x^2+1}\)

Vì \(x^2\ge0\)\(\Rightarrow x^2+1\ge1\)\(\Rightarrow\frac{3}{x^2+1}\le3\)\(\Rightarrow B\le3\)

Dấu " = " xảy ra \(\Leftrightarrow x^2=0\)\(\Leftrightarrow x=0\)( thoả mãn ĐKXĐ )

Vậy \(maxB=3\)\(\Leftrightarrow x=0\)

\(B=\frac{3\left(x+1\right)}{x^3+x^2+x+1}\)

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

Vì \(x^2\ge0\Rightarrow x^2+1\ge1\)

Mà \(\frac{3}{x^2+1}\le3\)Nên \(\Rightarrow B\le3\)

Dấu ''='' xảy ra <=> x = 0 

Vậy \(Max_B=3\Leftrightarrow x=0\)

\(B=\frac{3\left(x+1\right)}{x^3+x^2+x+1}=\frac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)

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

Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+1\ge1\forall x\)

\(\Rightarrow\frac{1}{x^2+1}\le1\forall x\)\(\Rightarrow\frac{3}{x^2+1}\le3\forall x\)

hay \(B\le3\)

Dấu " = " xảy ra \(\Leftrightarrow x^2=0\)\(\Leftrightarrow x=0\)

Vậy \(maxB=3\)\(\Leftrightarrow x=0\)

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

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

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)

\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)

b. -Để M thuộc Z thì:

\(\left(x^2+x-2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)

\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)

\(\Rightarrow4⋮\left(x+3\right)\)

\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)

c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)

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

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

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

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

\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)