\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)

\(P_1=\frac{3x^2+6x+10}{x^2+2x+3}\)

      \(=3+\frac{1}{x^2+2x+3}\)

Lại có: \(x^2+2x+3\)

          \(=\left(x+1\right)^2+2\ge2\)

\(\Rightarrow P_1\le3+\frac{1}{2}=\frac{7}{2}\)

Dấu = xảy ra khi x=-1

P₂ tương tự

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

Ta có :

\(2B=\frac{6x^2+12x+20}{x^2+2x+3}=\frac{7x^2+14x+21-x^2-2x-1}{x^2+2x+3}=\frac{7\left(x^2+2x+3\right)-\left(x+1\right)^2}{x^2+2x+3}\)

\(=7-\frac{\left(x+1\right)^2}{x^2+2x+3}\le7\) (Vì \(\frac{\left(x+1\right)^2}{x^2+2x+3}\ge0\))

Do \(2B\le7\Rightarrow B\le\frac{7}{2}\)đạt GTLN là \(\frac{7}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2+2x+3}=0\Rightarrow x=-1\)

Vậy GTLN của \(B\) là \(\frac{7}{2}\) tại \(x=-1\)

bài 1:

b, x² - 6x +10=x² - 2.x.3 +9 +1=(x - 3)² +1

Vì (x-3)²  >= 0 với mọi x

=> (x-3)² +1 >= 1 với mọi x

vậy GTNN của biểu thức bằng 1 <=> x-3=0<=> x=3

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a,\(M=-2x^2+2x-3\)

\(\Rightarrow2M=-4x^2+4x-6=-\left(4x^2-4x+1\right)-5=-\left(2x-1\right)^2-5\)

Vì\(-\left(2x-1\right)^2\le0\Rightarrow2M=-\left(2x-1\right)^2-5\le-5\Rightarrow M\le-\frac{5}{2}\)

Dấu "=" xảy ra khi x=1/2

Vậy Mmax=-5/2 khi x=1/2

b, \(N=3x-x^2-4=-x^2+3x-4=-\left(x^2-3x+\frac{9}{4}\right)-\frac{7}{4}=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\)

Vì \(-\left(x-\frac{3}{2}\right)^2\le0\Rightarrow N=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)

Dấu "=" xảy ra khi x=3/2

Vậy Nmax=-7/4 khi x=3/2

c, \(P=\frac{3}{x^2-6x+10}=\frac{3}{x^2-6x+9+1}=\frac{3}{\left(x-3\right)^2+1}\)

Vì \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1\Rightarrow\frac{1}{\left(x-3\right)^2+1}\le1\Rightarrow\frac{3}{\left(x-3\right)^2+1}\le3\)

Dấu "=" xảy ra khi x=3

Vậy Pmax=3 khi x=3

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

Ta có: \(A=\frac{3x^2+6x+11}{x^2+2x+3}=3+\frac{2}{x^2+2x+3}=3+\frac{2}{\left(x+1\right)^2+2}\)

Đặt \(B=\frac{2}{\left(x+1\right)^2+2}\),để A đạt giá trị lớn nhất thì B lớn nhất.

Mà B lớn nhất khi \(\left(x+1\right)^2+2\) bé nhất. 

Lại có: \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+2\ge2\) (1)

Từ (1) suy ra: \(B\le\frac{2}{2}=1\Rightarrow A=3+B\le3+1=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(A_{max}=4\Leftrightarrow x=-1\)