a; Đặt \(A=\frac{3x+7}{x^2+x+8}\)
=>\(A\left(x^2+x+8\right)=3x+7\)
=>\(A\cdot x^2+x\left(A-3\right)+8A-7=0\) (1)
\(\Delta=\left(A-3\right)^2-4A\left(8A-7\right)\)
\(=A^2-6A+9-32A^2+28A=-31A^2+22A+9\)
Để (1) có nghiệm thì \(-31A^2+22A+9\ge0\)
=>\(31A^2-22A-9\le0\)
=>\(31A^2-31A+9A-9\le0\)
=>(A-1)(31A+9)<=0
=>\(-\frac{9}{31}\le A\le1\)
=>\(A_{\min}=-\frac{9}{31}\) và \(A_{\max}=1\)
Đặt \(A=-\frac{9}{31}\)
=>\(\frac{3x+7}{x^2+x+8}=-\frac{9}{31}\)
=>\(-9\left(x^2+x+8\right)=31\left(3x+7\right)\)
=>\(9x^2+9x+72=-93x-217\)
=>\(9x^2+102x+289=0\)
=>\(\left(3x+17\right)^2=0\)
=>3x+17=0
=>3x=-17
=>\(x=-\frac{17}{3}\)
=>\(A_{\min}=-\frac{9}{31}\) khi \(x=-\frac{17}{3}\)
Đặt A=1
=>\(x^2+x+8=3x+7\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
=>\(A_{\max}=1\) khi x=1
b: Đặt \(B=\frac{2x-5}{x^2+4x+14}\)
=>\(B\left(x^2+4x+14\right)=2x-5\)
=>\(B\cdot x^2+x\left(4B-2\right)+14B+5=0\) (2)
\(\Delta=\left(4B-2\right)^2-4B\left(14B+5\right)\)
\(=16B^2-16B+4-56B^2-20B=-40B^2-36B+4=-4\left(10B^2+9B-1\right)\)
Để(2) có nghiệm thì Δ>=0
=>\(-4\left(10B^2+9B-1\right)\ge0\)
=>\(10B^2+9B-1\le0\)
=>\(10B^2+10B-B-1\le0\)
=>(B+1)(10B-1)<=0
=>\(-1\le B\le\frac{1}{10}\)
=>\(B_{\min}=-1;B_{\max}=\frac{1}{10}\)
Đặt B=-1
=>\(\frac{2x-5}{x^2+4x+14}=-1\)
=>\(x^2+4x+14=-2x+5\)
=>\(x^2+6x+9=0\)
=>\(\left(x+3\right)^2=0\)
=>x+3=0
=>x=-3
=>\(B_{min}=-1\) khi x=-3
Đặt \(B=\frac{1}{10}\)
=>\(\frac{2x-5}{x^2+4x+14}=\frac{1}{10}\)
=>\(x^2+4x+14=10\left(2x-5\right)\)
=>\(x^2+4x+14=20x-50\)
=>\(x^2-16x+64=0\)
=>\(\left(x-8\right)^2=0\)
=>x-8=0
=>x=8
=>\(B_{\max}=\frac{1}{10}\) khi x=8
