Giải:

a) Ta có: \(-x^2-6x+15\)

\(=-x^2-6x-9+24\)

\(=-\left(x^2+6x+9\right)+24\)

\(=-\left(x+3\right)^2+24\)

Vì \(-\left(x+3\right)^2\le0;\forall x\)

\(\Leftrightarrow-\left(x+3\right)^2+24\le24\)

Vậy giá trị lớn nhất của biểu thức trên là 24.

Câu b làm tương tự (phân tích đa thức thành nhân tử hoặc đưa về dạng hằng đẳng thức).

a)

\(E=-x^2-6x+15\)

\(E=-\left(x^2+6x-15\right)\)

\(E=-\left(x^2+2.x.3+9-24\right)\)

\(E=-\left(x+3\right)^2+24\)

Ta có: \(-\left(x+3\right)^2\le0\) với mọi x thuộc R

\(\Rightarrow-\left(x+3\right)^2+24\le24\)

Vậy GTLN của E = 24 khi x = -3

b)

\(F=\left(x^2+7x-2\right)\left(x^2+7x+2\right)\)

\(F=\left(x^2+7x\right)^2-2^2\)

\(F=\left(x^2+7x\right)^2-4\)

\(F=x^2\left(x+7\right)^2-4\)

Ta có:

\(x^2\left(x+7\right)^2\ge0\) với mọi x thuộc R

\(\Rightarrow x^2\left(x+7\right)^2-4\ge-4\) với mọi x thuộc R

Vậy GTNN của F = -4 khi x = 0 hoặc x = -7

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

a) \(A=\left\{{}\begin{matrix}x\ne-1;-2\\\dfrac{1}{x^2-x+1}\end{matrix}\right.\)

b)

\(A>1;\dfrac{1}{x^2-x+1}>1\Leftrightarrow x^2-x< 0\Leftrightarrow0< x< 1\)

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

x>0 => P >0 đang tìm Giá trị LN => chỉ xét P>0 <=> x>0

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

áp co si hai số dương x ; 1/x

\(\dfrac{1}{P}\ge2.\sqrt{x.\dfrac{1}{x}}+2=4\Rightarrow P\le\dfrac{1}{4}\)

đẳng thức khi x =1/x => x=1 thỏa mãn đk của x

\(MaxP=\dfrac{1}{4}\)

Câu 1:

\(A=\dfrac{81x}{3-x}+\dfrac{3}{x}=\dfrac{81x}{3-x}+\left(\dfrac{3}{x}-1\right)+1=\dfrac{81x}{3-x}+\dfrac{3-x}{x}+1\ge2\sqrt{\dfrac{81x}{3-x}.\dfrac{3-x}{x}}+1=18+1=19\)

Dấu "=" xảy ra <=> x = 0,3

Câu 2:

\(\dfrac{1}{3x-2\sqrt{6x}+5}=\dfrac{1}{\left(3x-2\sqrt{6x}+2\right)+3}=\dfrac{1}{\left(x\sqrt{3}-\sqrt{2}\right)^2+3}\le\dfrac{1}{3}\)

Dấu "=" xảy ra <=> \(x=\sqrt{\dfrac{2}{3}}\)

Câu 3:

\(A=2014\sqrt{x}+2015\sqrt{1-x}=2014\left(\sqrt{x}+\sqrt{1-x}\right)+\sqrt{1-x}\)

Ta có: \(\left(\sqrt{x}+\sqrt{1-x}\right)^2=x+1-x+2\sqrt{x\left(1-x\right)}=1+2\sqrt{x\left(1-x\right)}\ge1\)

=> \(A=2014\left(\sqrt{x}-\sqrt{1-x}\right)+\sqrt{1-x}\ge2014+\sqrt{1-x}\ge2014\)

Dấu "=" xảy ra <=> x = 1

a) \(A=\frac{3x^2+6x+10}{x^2+2x+3}\)

\(A=\frac{3x^2+6x+9+1}{x^2+2x+3}\)

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

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

\(A=3+\frac{1}{^{\left(x+1\right)^2+2}}\le3+\frac{1}{2}=\frac{7}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-1\)

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)

\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)

\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)

\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)

c: \(x^3-x^2+2=0\)

=>\(x^3+x^2-2x^2+2=0\)

=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)

=>x+1=0

=>x=-1

Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)