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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
\(P_1=\frac{3x^2+6x+10}{x^2+2x+3}\)
\(=3+\frac{1}{x^2+2x+3}\)
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
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 \(A_{max}=4\Leftrightarrow x=-1\)
\(P=\dfrac{3x^2+6x+11}{x^2+2x+3}\)
\(P=\dfrac{4x^2+8x+12-x^2-2x-1}{x^2+2x+3}\)
\(P=\dfrac{4\left(x^2+2x+3\right)}{x^2+2x+3}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\)
\(P=4-\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+2}\)
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 :
\(\frac{3x^2-6x+17}{x^2-2x+5}=3+\frac{2}{x^2-2x+5}\)
Biểu thức đạt giá trị lớn nhất
<=> x2 - 2x + 5 nhỏ nhất
Ta lại có :
x2 - 2x + 5 = x2 - 2x + 1 + 4 = (x - 1)2 + 4
Vì \(\left(x-1\right)^2\ge0\)
=> \(\left(x-1\right)^2+4\ge4\)
=> \(Min=4\)
Vậy giá trị lớn nhất của biểu thức là :
\(3+\frac{2}{4}=3+\frac{1}{2}=\frac{7}{2}\)
\(\frac{3x^2-6x+17}{x^2-2x+5}=\frac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}=3+\frac{2}{x^2-2x+5}=3+\frac{2}{\left(x-1\right)^2+4}\) (1)
Vì \(\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow\frac{2}{\left(x-1\right)^2+4}\le\frac{2}{4}=\frac{1}{2}\forall x\)
\(\Rightarrow3+\frac{2}{\left(x-1\right)^2+4}\le3+\frac{1}{2}=\frac{7}{2}\forall x\)
Dấu "=" xảy ra <=> \(x=1\)
Vậy ..........
\(M=-2x^2+2x-3\\ \Leftrightarrow2M=-4x^2+4x-6\\ \Leftrightarrow2M=-\left(4x^2-4x+4\right)-2\\ \Leftrightarrow2M=-\left(2x-2\right)^2-2\\ \Leftrightarrow M=\dfrac{-\left(2x-2\right)^2-2}{2}\)
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 :
\(\left(x-3\right)^2\ge0\\ \Rightarrow\left(x-3\right)^2+1\ge1\\ \Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\ge\dfrac{3}{1}\Rightarrow P\ge3\\ \Rightarrow Min\left(P\right)=3\Leftrightarrow x-3=0\Rightarrow x=3\)
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 \(x=-1\)
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
hay x<=4
b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)
=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)
=>12x+2+3x+9>=30x+18+48-20x
=>15x+11>=10x+66
=>5x>=55
hay x>=11
\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)
\(P_{max}=\dfrac{7}{2}\) 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 \(x=-\dfrac{3}{2}\)