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Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
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Ta có: 3x + y = 1 => y = 1 - 3x
a, Thay y = 1 - 3x vào M, ta có:
\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)
\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)
Vậy GTNN M = 1/4 khi x = y = 1/4
b, Thay y = 1 - 3x vào N
\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)
\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)
Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)
Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2
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Ta có: \(M=\frac{x^2+2x+3}{x^2+2}=\frac{2.\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}\)
\(=\frac{2.\left(x^2+2\right)}{x^2+2}-\frac{x^2-2x+1}{x^2+2}=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\)
Dấu "=" xảy ra khi \(x-1=0\Rightarrow x=1\)
Vậy Mmax = 2 khi x = 1
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2.
A = xy + 2yz + 3xz = xy + xz + 2yz + 2xz = x(y + z) + 2z(y + z)
Áp dụng BĐT: (a+b)^2/4 ≥ ab dấu = khi a = b
Ta có:
(x + y + z)^2/4 ≥ x(y + z)
(x+ y +z)^2/4 ≥ z(y + z)
=> A ≤ 3(x + y + z)^2/4 = 3.36/4 = 27
=> A max = 27 xảy ra khi:
{x = y + z
{z = y + z
<=> y = 0 và x = z = 3
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1)???
2) \(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=2+\dfrac{x^2-4x+4}{x^2-2x+1}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)
Vậy GTNN của A là 2 tại x=2.
3) \(\)Đặt \(a=\dfrac{1}{x+100}\Rightarrow x=\dfrac{1}{a}-100\)
\(D=\dfrac{x}{\left(x+100\right)^2}=a^2x=a^2\left(\dfrac{1}{a}-100\right)=a-100a^2=-100\left(a^2-\dfrac{a}{100}+\dfrac{1}{40000}-\dfrac{1}{40000}\right)=-100\left(a-\dfrac{1}{200}\right)^2+\dfrac{1}{400}\le\dfrac{1}{400}\)
Vậy GTLN của D là \(\dfrac{1}{400}\) tại \(a=\dfrac{1}{200}\Leftrightarrow x=100\)
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B= 6x+11/x^2-2x+3
= 9(x^2-2x+3)-9x^2+18x-27+6x+11/ x^2-2x+3
= 9 +
-(3x-4)^2/(x-1)^2+2
Vì (3x-4)^2 > hoặc = 0 với mọi x
=> -(3x-4)^2< hoặc =0
(x-1)^2+2>0 với mọi x
=> -(3x-4)^2/(x-1)^2+2< hoặc=0
=> B< hoặc =9
Vậy GTLN của B=9 khi x=4/3
Làm tương tự ta có gtnn của B=-1/2 khi x=-5
Chúc bạn học tốt!