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\(M=-2x^2+2x-3\\ \Leftrightarrow2M=-4x^2+4x-6\\ \Leftrightarrow2M=-\left(4x^2-4x+4\right)-2\\ \Leftrightarrow2M=-\left(2x-2\right)^2-2\\ \Leftrightarrow M=\dfrac{-\left(2x-2\right)^2-2}{2}\)
Ta có :
\(-\left(2x-2\right)^2\le0\\ \Rightarrow-\left(2x-2\right)^2-2\le-2\\ \Rightarrow\dfrac{-\left(2x-2\right)^2-2}{2}\le\dfrac{-2}{2}\\ \Rightarrow M\le-1\)
\(\Rightarrow Max\left(M\right)=-1\Leftrightarrow2x-2=0\Rightarrow x=1\)
.......
\(N=3x-x^2-4\\ \Leftrightarrow N=-\left(x^2-3x+4\right)\\ \Leftrightarrow N=-\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+\dfrac{16}{4}\right)\\ \Rightarrow N=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Ta có :
\(-\left(x-\dfrac{3}{2}\right)^2\le0\\ \Rightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\le0+\dfrac{7}{4}\\ \Rightarrow N\le\dfrac{7}{4}\\ \Rightarrow Max\left(M\right)=\dfrac{7}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(P=\dfrac{3}{x^2-6x+10}=\dfrac{3}{\left(x-3\right)^2+1}\)
Ta có :
\(\left(x-3\right)^2\ge0\\ \Rightarrow\left(x-3\right)^2+1\ge1\\ \Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\ge\dfrac{3}{1}\Rightarrow P\ge3\\ \Rightarrow Min\left(P\right)=3\Leftrightarrow x-3=0\Rightarrow x=3\)
\(P_1=\frac{3x^2+6x+10}{x^2+2x+3}\)
\(=3+\frac{1}{x^2+2x+3}\)
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
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
\(P=\dfrac{3x^2+6x+11}{x^2+2x+3}\)
\(P=\dfrac{4x^2+8x+12-x^2-2x-1}{x^2+2x+3}\)
\(P=\dfrac{4\left(x^2+2x+3\right)}{x^2+2x+3}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\)
\(P=4-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\)
Do : \(-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\) ≤ 0 ∀x
⇒ \(4-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\) ≤ 4
⇒ PMax = 4 ⇔ x = - 1
\(P=\dfrac{3x^2+6x+11}{x^2+2x+3}=\dfrac{3x^2+6x+9+2}{x^2+2x+3}=\dfrac{3\left(x^2+2x+3\right)+2}{x^2+2x+3}=3+\dfrac{2}{x^2+2x+3}=3+\dfrac{2}{\left(x+1\right)^2+2}\le3+1=4\)
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 \(A_{max}=4\Leftrightarrow x=-1\)
\(A=\dfrac{1}{-x^2+2x-2}\)
A min \(\Leftrightarrow\dfrac{1}{A}\)max
ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)
\(\dfrac{1}{A}\)max= -1 tại x=1
=> A min = -1 tại x=1
\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x2 nha bn)
B min \(\Leftrightarrow\dfrac{1}{B}\) max
ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)
\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2
\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2
\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)
A max \(\Leftrightarrow\dfrac{1}{A}\) min
\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)
\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)
\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)
B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)
B max \(\Leftrightarrow\dfrac{1}{B}\) min
\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)
\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)
=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)
Đây chỉ là gợi ý !! bn pải tự lí luận nha
tik 
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