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1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
1)
a) \(5x\left(x^2-3x+\dfrac{1}{5}\right)\)
\(=5x^3-15x^2+x\)
b) \(\left(x-3\right)\left(2x-1\right)\)
\(=2x^2-x-6x+3\)
\(=2x^2-7x+3\)
2)
a) \(3x^2-15xy\)
\(=3x\left(x-5y\right)\)
b) \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right)\left(x-3+y\right)\)
c) \(x^2+3x+2\)
\(=\left(x^2+x\right)+\left(2x+2\right)\)
\(=x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
bài 4
vì x2+1 >0 với mọi x , do đó GT của Q luôn xác định với mọi x
Q=\(\dfrac{2x^2-4x+5}{x^2+1}=\dfrac{\left(3x^2+3\right)+\left(2x^2-4x+2\right)}{x^2+1}\)=\(\dfrac{3\left(x^2+1\right)+2\left(x-1\right)^2}{x^2+1}=\dfrac{3\left(x^2+1\right)}{x^2+1}+\dfrac{2\left(x-1\right)^2}{x^2+1}\)=\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\)
Do (x-1)2 ≥ 0
=>2(x-1)2 ≥ 0
x2+1 ≥ 0
=>\(\dfrac{2\left(x-1\right)^2}{x^2+1}\ge0\)
=>\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\ge3\)
=> Q ≥ 3
=>GTNN của Q =3 khi
x-1=0
=>x=1
Vậy GTNN của Q =3 khi x=1
a)
2x-3=0 => x=3/2
b)
2x^2 +1 =0 => vô nghiệm
c) x^2 -25 =0 => x=5 loiaj
x=-5 nhân
d)
x^2 -25 =0 => x=5 loại
x=-5 loại
a)
\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)
b)
x khác 1
c)
x khác 0; x khác 5
d) x khác 5 ; x khác -5