M xác định
\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3}{x-1}+\frac{1}{x\left(x-1\right)}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}=\frac{3x+1}{x\left(x-1\right)}\)
Thay x=5 ta có:
\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)
Vậy \(M=5\)tại x=5
\(M=0\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)
Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)
\(M=-1\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy với \(x=-1\)thì \(M=-1\)
\(4x^2+4x+11\)
\(=\left(2x\right)^2+2.2x.1+1^2+10\)
\(=\left(2x+1\right)^2+10\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2+10\ge10\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy Min \(4x^2+4x+11=10\Leftrightarrow x=-\frac{1}{2}\)
a, Để M được xác định thì \(\frac{3}{x-1}\)và \(\frac{1}{x^2-x}\)xác định
=> \(\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x-1\ne0\\x\left(x-1\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}}\)
Vậy:...
b, Ta có:
\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}\)
\(=\frac{3x+1}{x\left(x+1\right)}\)
+) Với x = 5
\(\Rightarrow M=\frac{3.5+1}{5.\left(5+1\right)}=\frac{8}{15}\)
c, Nếu M = 0
\(\Rightarrow\)\(\frac{3x+1}{x\left(x+1\right)}=0\)
\(\Rightarrow3x+1=0\) ( Vì M xác định nên x ( x + 1 ) khác 0 )
=> 3x = -1
=> x = \(-\frac{1}{3}\)
Vậy:...
d, Nếu M = -1
\(\Rightarrow\frac{3x+1}{x\left(x+1\right)}=-1\)
\(\Rightarrow3x+1=-x\left(x+1\right)\)
\(\Rightarrow3x+1=-x^2-x\)
\(\Rightarrow x^2+4x+1=0\)
\(\Rightarrow x^2+2.x.2+2^2-3=0\)
\(\Rightarrow\left(x+2\right)^2=3\)
\(\Rightarrow\orbr{\begin{cases}x+2=-\sqrt{3}\\x+2=\sqrt{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\sqrt{3}-2\\x=\sqrt{3}-2\end{cases}}\)
Vậy:....
(2) Ta có: 4x2 +4x + 11
= (2x)2 + 2.2x.1 + 12 + 10
= ( 2x + 1 )2 + 10
Vì: ( 2x + 1 )2 \(\ge\)0 với mọi x
=> ( 2x + 1 )2 + 10 \(\ge\)10
Dấu "=" xảy ra khi và chỉ khi: 2x + 1 = 0
=> 2x = -1
=> x = -1/2
Vậy: GTNN 4x2 + 4x + 11 là 10 khi vs chỉ khi x = -1/2
\(1,a.ĐKXĐ:\hept{\begin{cases}x\ne0\\x-1\ne0\end{cases}\Leftrightarrow x\ne0;1}\)
\(b.M=\frac{3}{x-1}+\frac{1}{x^2-x}\)
\(=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}\)
\(=\frac{3x+1}{x\left(x-1\right)}\)
Với x = 5 thỏa mãn ĐKXĐ
khi đó \(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{16}{20}=\frac{4}{5}\)
\(c.M=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow x=-\frac{1}{3}\left(TmĐKXĐ\right)\)
\(d.M=-1\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\)
\(\Leftrightarrow3x+1=-x\left(x-1\right)\)
\(\Leftrightarrow3x+1=-x^2+x\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\left(TmĐKXĐ\right)\)
\(2,4x^2+4x+11=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy ........
