Tìm GTLN: a) —x^2—4x+9
b) x(6—x)+74+x
c) 5x—x^2
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a) Q=13-(x^2+4x+4)=13-(x+2)^2<=13 Qmax=13 khi x=-2
b) M=\(6x-x^2+74+x=74-\left(x^2+7x\right)=74-\left(x^2-2.\frac{7}{2}x+\left(\frac{7}{2}\right)^2\right)^{^2}-\left(\frac{7}{2}\right)^2\\ \)
\(\frac{74\cdot4-49}{4}-\left(x-\frac{7}{2}\right)^2\le\frac{74\cdot4-49}{4}=M_{max}\)đảng thức khi x=7/2
C) \(P=\frac{25}{4}-\left(x^2-2.\frac{5}{2}+\left(\frac{5}{2}\right)^2\right)=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}=P_{max}\) khi x=5/2
\(M=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{6}{x+2}\right)\)
a) dkxd : x khac {0;1;-2)
\(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right).\left(\dfrac{x+2}{6}\right)\)
\(M=\left(\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(\dfrac{x+2}{6}\right)=\dfrac{-6}{6\left(x-2\right)}=\dfrac{1}{2-x}\)
b)
GTLN M =1 khi x =1
a: \(A=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=5/2
b: \(B=x^2-4x+4+y^2-8y+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu '=' xảy ra khi x=2 và y=4
a: =-x^2+6x-4
=-(x^2-6x+4)
=-(x^2-6x+9-5)
=-(x-3)^2+5<=5
Dấu = xảy ra khi x=3
b: =3(x^2-5/3x+7/3)
=3(x^2-2*x*5/6+25/36+59/36)
=3(x-5/6)^2+59/12>=59/12
Dấu = xảy ra khi x=5/6
c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)
\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)
\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)
Dấu = xảy ra khi x=4 hoặc x=2
2) a) Đặt \(\left(x-1\right)\left(x+6\right)=t\)
\(\Leftrightarrow x^2+5x-6=t\)
\(\left(x+2\right)\left(x+3\right)=x^2+5x+6=t+12\)
\(A=t\left(t+12\right)+2042\)
\(A=t^2+12t+2042\)
\(A=\left(t+6\right)^2-6^2+2042\)
\(A=\left(t+6\right)^2+2006\)
\(\left(t+6\right)^2\ge0\Rightarrow\left(t+6\right)^2+2006\ge2006\)
\(Min_A=2006\) khi \(\left(t+6\right)^2=0\Leftrightarrow t=-6\Leftrightarrow x^2+5x-6=-6\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy: MinA=2006 khi x=0 hoặc x=5
Bài 2b làm tương tự
\(a,x^2+5y^2+2xy-4x-8y+2015\)
\(=\left(x^2+y^2+2xy\right)-4\left(x+2y\right)+4+4y^2-4y+1+2015=\left[\left(x+y\right)^2-4\left(x+2y\right)+4\right]+\left(4y^2-4y+1\right)+2015\)
\(=\left(x+y-2\right)^2+\left(2y-1\right)^2+2010\)
Do.....
Nên .....
Vậy MIN = 2010 <=> x = 3/2; y = 1/2
P/S: nhương người đi sau
\(\)
a) \(A=-x^2-4x+9=-\left(x^2+2.x.2+2^2\right)+13\)
\(=-\left(x+2\right)^2+13\)
Ta có: \(\left(x+2\right)^2\ge0\)
\(\Rightarrow-\left(x+2\right)^2\le0\)
\(\Rightarrow-\left(x+2\right)^2+13\le13\)
Dấu ''='' xảy ra khi \(x=-2\)
Vậy GTLN của A bằng 13 đạt được khi \(x=-2\)
b) \(B=x\left(6-x\right)+74+x=-x^2+7x+74\)
\(=-\left[x^2-2\times x\times\frac{7}{2}+\left(\frac{7}{2}\right)^2\right]+86,25\)
\(=-\left(x-\frac{7}{2}\right)^2+86,25\)
Ta có: \(\left(x+\frac{7}{2}\right)^2\ge0\)
\(\Rightarrow-\left(x+\frac{7}{2}\right)\le0\)
\(\Rightarrow-\left(x+\frac{7}{2}\right)^2+86,25\le86,25\)
Dấu '=' xảy ra khi \(x=-\frac{7}{2}=-3,5\)
Vậy GTLN của B bằng 86,25 đạt được khi x = -3,5
c) \(C=5x-x^2=-\left[x^2-2\times x\times\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]+6,25\)
\(=-\left(x-\frac{5}{2}\right)^2+6,25\)
Ta có: \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow-\left(x-\frac{5}{2}\right)^2\le0\)
\(\Rightarrow-\left(x-\frac{5}{2}\right)^2+6,25\le6,25\)
Dấu '' = '' xảy ra khi \(x=\frac{5}{2}=2,5\)
Vậy GTLN của C bằng 6,25 đạt được khi x = 2,5