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\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{6}=1+\dfrac{x-2}{6}\)
\(=\dfrac{6}{6}+\dfrac{x-2}{6}=\dfrac{x+4}{6}\)
b/ +) \(P=0\Leftrightarrow\dfrac{x+4}{6}=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\) (tm)
Vậy x = -4 thì P = 0
+) \(P=1\Leftrightarrow\dfrac{x+4}{6}=1\Leftrightarrow x+4=6\Leftrightarrow x=2\) (ktm)
K có gt nào của x tm P = 1
c/ \(P>0\Leftrightarrow\dfrac{x+4}{6}>0\Leftrightarrow x+4>0\Leftrightarrow x>-4\)
\(\forall x\ne\pm2;x\ne-3\)
\(B=\dfrac{16x^2+4x+1}{2x}=\dfrac{2x\left(8x+2\right)+1}{2x}=8x+2+\dfrac{1}{2x}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(8x+\dfrac{1}{2x}\) ≥ \(2\sqrt{8x.\dfrac{1}{2x}}=2\sqrt{4}=4\)
⇔ \(8x+\dfrac{1}{2x}\) + 2 ≥ 4 + 2 = 6
⇒ \(B_{Min}=6\) ⇔\(8x=\dfrac{1}{2x}\) ⇔ \(x=\dfrac{1}{4}\)
\(a.P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}.\dfrac{\left(x+2\right)\left(x-2\right)}{2x+4-3-x+2}=\left(x+3\right).\dfrac{x-2}{x+3}=x-2\left(x\ne\pm2;x\ne-3\right)\)
\(b.P=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(KTM\right)\)
\(P=1\Leftrightarrow x-2=1\Leftrightarrow x=3\left(TM\right)\)
\(c.P>0\Leftrightarrow x-2>0\Leftrightarrow x>2\)
\(P=-\left(4x^2-4x+1+x+\frac{1}{4x}-2015\right)\)
\(=-\left[\left(2x-1\right)^2+\frac{\left(2x-1\right)^2}{4x}\right]+2014\)
\(P\le2014\forall x>0\)
Dấu "=" xảy ra <=> x=\(\frac{1}{2}\)
a,\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{-3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{\left(x+2\right)}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}:\dfrac{6}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{6}=\dfrac{x-2}{6}\)
b, Để P = 0 ⇔ \(\dfrac{x-2}{6}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Để \(P=1\Leftrightarrow\dfrac{x-2}{6}=1\Leftrightarrow x-2=6\Leftrightarrow x=8\)
c, Để P > 0 \(\Leftrightarrow\dfrac{x-2}{6}>0\Leftrightarrow x-2>6\Leftrightarrow x>8\)
\(A=\dfrac{1}{-x^2+2x-2}\)
A min \(\Leftrightarrow\dfrac{1}{A}\)max
ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)
\(\dfrac{1}{A}\)max= -1 tại x=1
=> A min = -1 tại x=1
\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x2 nha bn)
B min \(\Leftrightarrow\dfrac{1}{B}\) max
ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)
\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2
\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2
\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)
A max \(\Leftrightarrow\dfrac{1}{A}\) min
\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)
\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)
\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)
B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)
B max \(\Leftrightarrow\dfrac{1}{B}\) min
\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)
\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)
=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)
Đây chỉ là gợi ý !! bn pải tự lí luận nha
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