1. \(M=\left(\frac{1}{a}+\frac{a}{a+1}\right):\frac{a}{a^2+a}\) \(\left(a\ne0,a\ne-1\right)\)
\(=\frac{a+1+a^2}{a\left(a+1\right)}.\frac{a\left(a+1\right)}{a}=\frac{a^2+a+1}{a}\)
\(2.\left(a-5\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-5=0\\a+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=5\left(t/m\right)\\a=-1\left(loai\right)\end{matrix}\right.\)
Với a = 5 thì \(M=\frac{5^2+5+1}{5}=\frac{31}{5}\)
3. \(M=\frac{a^2+a+1}{a}=a+1+\frac{1}{a}\)
a> 0 => 1/a >0
Áp dụng BĐT cô si với hai số a và \(\frac{1}{a}\) , có:
\(a+\frac{1}{a}\ge2\sqrt{a.\frac{1}{a}}=2\)
\(\Leftrightarrow a+1+\frac{1}{a}\ge3\)
Dấu = xảy ra \(\Leftrightarrow a=\frac{1}{a}\Leftrightarrow a^2=1\Leftrightarrow a=1\) ( vì a > 0)
Vậy \(Min_M=3\Leftrightarrow a=1\)
1. Rút gọn biểu thức M.
\(M=\left(\frac{1}{a}+\frac{a}{a}+1\right):\frac{a}{a^2}+a\)
\(M=\left(\frac{1}{a}+1+1\right):\frac{1}{a}+a\)
\(M=\left(\frac{1}{a}+2\right).a+a\)
\(M=\left(\frac{1}{a}+\frac{2a}{a}\right).a+a\)
\(M=\frac{2a+1}{a}.a+a\)
\(M=\frac{a.\left(2a+1\right)}{a}+a\)
\(M=2a+1+a\)
\(M=3a+1\)
( HS 6->7 làm bài )
