tim max B=\(\frac{3x+5}{4x+1}\)
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1) A = 3 - 4x2 - 4x = - (4x2 + 4x +1) + 4 = - (2x+1)2 + 4
Vì - (2x+1)2 \(\le\)0 nên A = - (2x+1)2 + 4 \(\le\) 4 vậy maxA = 4 khi 2x+1 = 0 => x = -1/2
b) ta có x2 + 6x + 11 = x2 + 2.3x + 9 + 2 = (x+3)2 + 2 \(\ge\) 0 + 4 = 4
=> \(B=\frac{1}{x^2+6x+11}\le\frac{1}{4}\) vậy maxB = 1/4 khi x = -3
2) a) 3x2 - 3x + 1 = 3.(x2 - x) + 1 = 3.(x2 - 2.x\(\frac{1}{2}\) + \(\frac{1}{4}\)) + \(\frac{1}{4}\) = 3.(x - \(\frac{1}{2}\) )2 + \(\frac{1}{4}\) \(\ge\)0 + \(\frac{1}{4}\)= \(\frac{1}{4}\)
vậy min(3x2 - 3x + 1) = 1/4 khi x = 1/2
b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b| \(\ge\) |a - b|. dấu = khi a.b < 0
ta có: |3x - 3| + |3x - 5| \(\ge\) |3x - 3 - (3x - 5)| = |2| = 2
vậy min = 2 khi (3x - 3)(3x - 5) < 0 hay 1< x < 5/3
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\(M=\frac{3x^2-4x}{^{\left(x-1\right)^2}}=\frac{3\left(x^2-2x+1\right)+2\left(x-1\right)-1}{\left(x-1\right)^2}=3+\frac{2y-1}{y^2}\)
\(4-\left(\frac{1}{y^2}-\frac{2}{y}+1\right)=4-\left(\frac{1}{y}-1\right)^2\)
Mmax =4 khi y=1; x=2
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a, \(A=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)
=> \(A\ge-\frac{9}{4}\) dấu = xảy ra khi : \(x=\frac{-1}{2}\)
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Ta có \(A=\frac{4x-3x^2}{x^2+1}\)\(\Rightarrow\)A-1=\(\frac{4x-3x^2}{x^2+1}-1\)=\(\frac{4x-4x^2-1}{x^2+1}\)=\(\frac{-\left(4x^2-4x+1\right)}{x^2+1}=\frac{-\left(2x-1\right)^2}{x^2+1}\)\(\le0\)
dấu''='' xảy ra \(\Leftrightarrow-\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
vạy max A-1=0khi x=1/2 suy ra max A =1 khi x=1/2
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a: \(\Leftrightarrow9x^2-12x+4-6x^2-16x=0\)
\(\Leftrightarrow3x^2-28x+4=0\)
\(\text{Δ}=\left(-28\right)^2-4\cdot3\cdot4=736>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{28-4\sqrt{46}}{6}=\dfrac{14-2\sqrt{46}}{3}\\x_2=\dfrac{14+2\sqrt{46}}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow16x^2+24x+9-16x^2+25=12\)
=>24x+34=12
=>24x=-22
hay x=-11/12
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