tìm GTLN và GTNN
\(A=\dfrac{x^2+10x+16}{x^2+2x+2}\)
a, Tìm GTNN: A = \(\dfrac{x^2-2x+2013}{x^2}\) ; x>0
b, Tìm GTLN và GTNN của: B = \(\dfrac{4x+1}{4x^2+2}\)
a.
\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)
Dấu "=" xảy ra khi \(x=2013\)
b.
\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)
\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)
\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)
\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)
Tìm GTLN của Q=\(-2x^2+6x+8\)
Tìm GTLN và GTNN của: A=\(\dfrac{6x+17}{x^2+2}\)
\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)
\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)
\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)
\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)
\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)
\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)
Tìm GTNN và GTLN nếu có của các biểu thức
\(A=\dfrac{2x^2-2x+5}{\left(x+1\right)^2}\)
\(B=\dfrac{4x^2+x+4}{x^2+x+1}\)
Tìm GTNN hoặc GTLN :
a,x^2+y^2-2x+6y-30
b,4-2x^2
c,-x^2+10x-5
d,-3x^2+2x-5
giup mk với mk cần gấp lắm..rất gấp T^T
a, \(x^2+y^2-2x+6y-30\)
\(=x^2-2x+1+y^2+6y+9-40\)
\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)
\(min=-40\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
a)x^2+y^2-2x+6y-30=(x-1)^2+(y+3)^2-40\(\ge\) -40
dấu = xảy ra khi x=1,y=-3
b, \(4-2x^2\le4\)
\(max=4\Leftrightarrow x=0\)
c, \(-x^2+10x-5=-\left(x^2-10x+25\right)+20=-\left(x-5\right)^2+20\le2\text{}0\)
\(max=20\Leftrightarrow x=5\)
tìm gtnn hoặc gtln
a, A=-6x+x^2+11
b,B=-1+2x^x+10x
a) Ta có : \(A=-6x+x^2+11\)
\(\Rightarrow A=\left(x^2-6x+9\right)+2\)
\(\Rightarrow A=\left(x-3\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(minA=2\Leftrightarrow x=3\)
b) \(B=-1+2x^x+10x\)
\(\Rightarrow\)Tớ đang thắc mắc cái chỗ 2xx :)))
\(\dfrac{1}{2x-x^2-4}\) tìm GTLN/ GTNN
\(\dfrac{3x^2+14}{x^2+4}\)
\(\dfrac{2x+1}{x^2+2}\)
Tìm GTLN GTNN của A=\(\dfrac{\text{ 2x+1}}{\text{x^2+2 }}\)
\(A=\dfrac{2x+1}{x^2+2}\)
\(\Leftrightarrow Ax^{2\:}+2A=2x+1\)
+) \(A=0\Rightarrow x=-\dfrac{1}{2}\)
+) \(A\ne0\)
\(Ax^2+2A=2x+1\)
\(\Leftrightarrow Ax^{2\:}-2x=1-2A\)
\(\Leftrightarrow x^2-2.\dfrac{x}{A}=\dfrac{1-2A}{A}\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{A}+\dfrac{1}{A^2}=\dfrac{1-2A}{A}+\dfrac{1}{A^2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{A-2A^2+1}{A^2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{\left(1-A\right)\left(2A+1\right)}{A^2}\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{A}\right)^2\ge0\left(\forall x,A\ne0\right)\\A^2\ge0\end{matrix}\right.\)
⇒ \(\left(1-A\right)\left(2A+1\right)\ge0\)
⇒ \(-\dfrac{1}{2}\le A\le1\)
Còn lại tụ làm nha
\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2-2+2x+1}{x^2+2}\\ =1-\dfrac{-\left(x-1\right)^2}{x^2+2}\\ Do\left(x-1\right)^2\ge0\Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}\ge0\\ \Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}=0\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}+1\le1\)
\(Dấu"="\Leftrightarrow A=1\\ \Leftrightarrow x-1=0\Rightarrow x=1\\ Vậy.P_{max}=1.khi.x=1\\ A=\dfrac{2x+1}{x^2+2}\rightarrow2A+1=\dfrac{2.\left(2x+1\right)}{x^2+2}+1\\ =\dfrac{4x+2+x^2+2}{x^2+2}=\dfrac{x^2+4x+2}{x^2+2}=\dfrac{\left(x+2\right)^2}{x^2+2}\\ Do\left(x+2\right)^2\ge0\Leftrightarrow\dfrac{\left(x+2\right)^2}{x^2+2}\ge0\)
\(Dấu"="\Leftrightarrow A=\dfrac{1}{2}khi.x=-2\\ \Rightarrow2A+1\ge0\Rightarrow2A\ge-1\Rightarrow A>-\dfrac{1}{2}\\ Vậy.MinA=-\dfrac{1}{2}.khi.x=-2\)
tìm GTNN hoặc GTLN
A=2x2 +10x-1
B=5x2-x
\(A=2x^2+10x-1=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)
=> Min A \(=-\frac{27}{2}\Leftrightarrow x=-\frac{5}{2}\)
\(B=5x^2-x=5\left(x-\frac{1}{10}\right)^2-\frac{1}{20}\ge-\frac{1}{20}\)
=> Min B \(=-\frac{1}{20}\Leftrightarrow x=\frac{1}{10}\)
Chứng minh đa thức sau vô nghiệm
2x^2+12x+19
Tìm gtnn
A) 4x2+4x+2
B) 3x2-30x+73
Tìm gtln
A) -x2-10x-34
B) -2x2-16-31
Đặt \(f\left(x\right)=-x^2-2x-3\)
\(=-x^2-x-x-3\)
\(=-x.\left(x-1\right)-\left(x-1\right)-2\)
\(=-[-\left(x-1\right)^2]-2\le-2< 0\)
\(\Rightarrow\)Đa thức không có nghiệm
Đặt \(A=-x^2-2x-3\)
\(\Rightarrow-A=x^2+2x+3\)
\(-A=\left(x^2+2x+1\right)+2\)
\(-A=\left(x+1\right)^2+2\)
\(\Rightarrow A=-\left(x+1\right)^2-2\)
Ta có: \(-\left(x+1\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+1\right)^2-2\le2\forall x\)
\(\Rightarrow\) Đa thức vô nghiệm
* Tìm GTNN :
\(a)\) Đặt \(A=4x^2+4x+2\) ta có :
\(A=\left(4x^2+4x+1\right)+1\)
\(A=\left[\left(2x\right)^2+2.2x.1+1^2\right]+1\)
\(A=\left(2x+1\right)^2+1\ge1\)
Dấu "=" xay ra \(\Leftrightarrow\)\(\left(2x+1\right)^2=0\)
\(\Leftrightarrow\)\(2x+1=0\)
\(\Leftrightarrow\)\(2x=-1\)
\(\Leftrightarrow\)\(x=\frac{-1}{2}\)
Vậy GTNN của \(A\) là \(1\) khi \(x=\frac{-1}{2}\)
Bài b) là tìm GTLN thì đúng hơn
\(b)\) Đặt \(B=-2x^2-16x-31\) ( hình như đề thiếu chỗ \(16x\) ) ta có :
\(-2B=4x^2+32x+62\)
\(-2B=\left(4x^2+32x+64\right)-2\)
\(-2B=\left[\left(2x\right)^2+2.2x.8+64\right]-2\)
\(-2B=\left(2x+8\right)^2-2\ge-2\)
\(B\le\frac{-2}{-2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(2x+8\right)^2=0\)
\(\Leftrightarrow\)\(2x+8=0\)
\(\Leftrightarrow\)\(2x=-8\)
\(\Leftrightarrow\)\(x=\frac{-8}{2}\)
\(\Leftrightarrow\)\(x=-4\)
Vậy GTLN của \(B\) là \(1\) khi \(x=-4\)
Chúc bạn học tốt ~