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Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
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Ta có: \(-x^2+4x+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\forall x\in R\)
⇒ max y = 7 tại x = 2
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\(A=\dfrac{4x+3}{x^2+1}\Leftrightarrow Ax^2+A=4x+3\\ \Leftrightarrow Ax^2-4x+A-3=0\)
Coi đây là PT bậc 2 ẩn x thì PT có nghiệm
\(\Leftrightarrow\Delta=16-4A\left(A-3\right)\ge0\\ \Leftrightarrow16-4A^2+12A\ge0\\ \Leftrightarrow-A^2+3A+4\ge0\\ \Leftrightarrow-1\le A\le4\)
Vậy \(A_{max}=4;A_{min}=-1\)
\(A_{max}=4\Leftrightarrow\dfrac{4x+3}{x^2+1}=4\Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ A_{min}=-1\Leftrightarrow\dfrac{4x+3}{x^2+1}=-1\Leftrightarrow x^2+1=-4x-3\Leftrightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)
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1:
a: =x^2-7x+49/4-5/4
=(x-7/2)^2-5/4>=-5/4
Dấu = xảy ra khi x=7/2
b: =x^2+x+1/4-13/4
=(x+1/2)^2-13/4>=-13/4
Dấu = xảy ra khi x=-1/2
e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4
Dấu = xảy ra khi x=1/2
f: x^2-4x+7
=x^2-4x+4+3
=(x-2)^2+3>=3
Dấu = xảy ra khi x=2
2:
a: A=2x^2+4x+9
=2x^2+4x+2+7
=2(x^2+2x+1)+7
=2(x+1)^2+7>=7
Dấu = xảy ra khi x=-1
b: x^2+2x+4
=x^2+2x+1+3
=(x+1)^2+3>=3
Dấu = xảy ra khi x=-1
-x^2+4x+3=-(x^2-4x-3)
=-(x^2-4x+2^2-4-3)
=-(x-2)^2+7
GTLN=7 tại x=2