7: \(y=\frac{x^2+3}{x-1}\)
\(=\frac{x^2-1+4}{x-1}=x+1+\frac{4}{x-1}\)
=>f'(x)=\(1+\left(\frac{4}{x-1}\right)^{\prime}=1+\frac{4^{\prime}\left(x-1\right)-4\left(x-1\right)^{\prime}}{\left(x-1\right)^2}=1-\frac{4}{\left(x-1\right)^2}\)
Đặt f'(x)=0
=>\(1-\frac{4}{\left(x-1\right)^2}=0\)
=>\(\frac{4}{\left(x-1\right)^2}=1\)
=>\(\left(x-1\right)^2=4\)
=>x-1=2 hoặc x-1=-2
=>x=3(nhận) hoặc x=-1(loại)
\(f\left(2\right)=\frac{2^2+3}{2-1}=7\)
\(f\left(4\right)=\frac{4^2+3}{4-1}=\frac{19}{3}\)
\(f\left(3\right)=\frac{3^2+3}{3-1}=\frac{9+3}{2}=\frac{12}{2}=6\)
=>\(f\left(x\right)_{\min\left\lbrack2;4\right\rbrack}\) =6 khi x=3
8: \(y=\frac{2x^2+5x+8}{x+8}\)
=>y'=\(\frac{\left(2x^2+5x+8\right)^{\prime}\left(x+8\right)-\left(x+8\right)^{\prime}\left(2x^2+5x+8\right)}{\left(x+8\right)^2}\)
=>y'=\(\frac{\left(4x+5\right)\left(x+8\right)-\left(2x^2+5x+8\right)}{\left(x+8\right)^2}=\frac{4x^2+37x+40-2x^2-5x-8}{\left(x+8\right)^2}\)
=>y'=\(\frac{2x^2+32x+32}{\left(x+8\right)^2}=\frac{2\left(x^2+16x+16\right)}{\left(x+8\right)^2}\)
Đặt y'=0
=>\(x^2+16x+16=0\)
=>\(x^2+16x+64=48\)
=>\(\left(x+4\right)^2=48\)
=>\(\left[\begin{array}{l}x+4=4\sqrt3\\ x+4=-4\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\sqrt3-4\left(loại\right)\\ x=-4\sqrt3-4\left(loại\right)\end{array}\right.\)
\(f\left(0\right)=\frac{2\cdot0^2+5\cdot0+8}{0+8}=1\)
\(f\left(8\right)=\frac{2\cdot8^2+5\cdot8+8}{8+8}=\frac{2\cdot64+48}{16}=2\cdot4+3=8+3=11\)
=>\(f\left(x\right)_{\max\left\lbrack0;8\right\rbrack}\) =11 khi x=8
