a/ĐKXĐ : \(2x^2-5x+2\ne0\Leftrightarrow\)\(\left(2x-1\right)\left(x-2\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(A=\dfrac{\left(2x\right)^2-2.2x.1+1}{2x^2-4x-x+2}=\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)\left(x-2\right)}=\dfrac{2x-1}{x-2}\)
b/ Thay x = 1/2 vào BT (A),ta có :
\(A=\dfrac{2.\dfrac{1}{2}-1}{\dfrac{1}{2}-2}=0\)
Thay x =3 vào biểu thức A,ta có :
\(A=\dfrac{2.3-1}{3-2}=5\)
c/ Để \(A=\dfrac{1}{2}\Leftrightarrow\dfrac{2x-1}{x-3}=\dfrac{1}{2}\Leftrightarrow4x-2=x-3\Leftrightarrow x=-\dfrac{1}{3}\)
d/\(A=\dfrac{2x-1}{x-2}=2+\dfrac{1}{x-2}\)
Để A có giá trị nguyên thì \(x-2\inƯ\left(1\right)\)
=> x- 2 = 1 => x = 3 < TM>
=> x-2 = -1 => x = 1 <TM>
Vậy ....
a) ĐK \(2x^2-5x+2\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(A=\dfrac{4x^2-4x+1}{2x^2-5x+2}=\dfrac{\left(2x\right)^2-2.2x.1+1^2}{2x^2-4x-x+2}\)
\(=\dfrac{\left(2x-1\right)^2}{2x\left(x-2\right)-\left(x-2\right)}=\dfrac{\left(2x-1\right)^2}{\left(x-2\right)\left(2x-1\right)}=\dfrac{2x-1}{x-2}\)
b) cho \(x=\dfrac{1}{2}\)
\(A=\dfrac{2x-1}{x-2}=\dfrac{2.\dfrac{1}{2}-1}{\dfrac{1}{2}-2}=0\)
cho x=3
\(A=\dfrac{2x-1}{x-2}=\dfrac{2.3-1}{3-2}=5\)
c) cho \(A=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2x-1}{x-2}=\dfrac{1}{2}\Leftrightarrow2\left(2x-1\right)=x-2\)
\(\Leftrightarrow4x-2=x-2\Leftrightarrow x=0\)
d) \(A=\dfrac{2x-1}{x-2}=\dfrac{2\left(x-2\right)+3}{x-2}=2+\dfrac{3}{x-2}\)
Để A nguyên khi \(\left(x-2\right)⋮3\)
vậy \(\left(x-2\right)\inƯ\left(3\right)=\left\{-1;1;3;-3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=-1\\x-2=1\\x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=5\\x=-1\end{matrix}\right.\)
