B1.\(\dfrac{x^2-4x+3}{x^2+x-2}\)(rút gọn phân thức)
B2.Tính
\(\left(\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\right):\dfrac{10}{4x+2y}\)
B3.
A=\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)và B=\(\dfrac{1}{x-1}\)
a.Tính giá trị của B khi \(x^2-x=0\)
b.Rút gọn b/thức A
c.Cho C=A-B.Tìm x để C =-2
d.Tìm giá trị nguyên của x để C có giá trị nguyên
Bài 3:
a: ĐKXĐ: \(x\notin\left\{1\right\}\)
\(x^2-x=0\)
=>x(x-1)=0
=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào B, ta được:
\(B=\dfrac{1}{0-1}=\dfrac{1}{-1}=-1\)
b: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-x-8}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x-4}{x-2}\)
c: C=A-B
\(=\dfrac{x-4}{x-2}-\dfrac{1}{x-1}\)
\(=\dfrac{\left(x-4\right)\left(x-1\right)-x+2}{\left(x-2\right)\left(x-1\right)}\)
\(=\dfrac{x^2-5x+4-x+2}{\left(x-2\right)\left(x-1\right)}=\dfrac{x^2-6x+6}{\left(x-2\right)\left(x-1\right)}\)
Để C=-2 thì \(-2\left(x-2\right)\left(x-1\right)=x^2-6x+6\)
=>\(-2\left(x^2-3x+2\right)-x^2+6x-6=0\)
=>\(-2x^2+6x-4-x^2+6x-6=0\)
=>\(-3x^2+12x-10=0\)
=>\(x=\dfrac{6\pm\sqrt{6}}{3}\)
Bài 1:
ĐKXĐ: \(x\notin\left\{-2;1\right\}\)
\(\dfrac{x^2-4x+3}{x^2+x-2}\)
\(=\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+2\right)\left(x-1\right)}\)
\(=\dfrac{x-3}{x+2}\)
em c.ơn trước