Bài 1 : Tìm x biết
a/ x ( x + 4 ) + x + 4=0
b/ x ( x - 3) + 2x - 6 = 0
Bài 2 : rút gọn biểu thức
a/ \(\dfrac{6x^2y^2}{8xy^5}\) b/ \(\dfrac{3x^2-x}{9x^2-6x+1}\) e/ \(\dfrac{x^2+7x+12}{x^2+5x+6}\)
c/ \(\dfrac{x^2-9}{x^2+6x+9}\) d/ \(\dfrac{x^2+2x+1}{3x+3}\)
Bài 3 : thực hiện phép tính ( các mẫu thức đều không buông )
a/ \(\dfrac{15}{2x+6}+\dfrac{5x}{2x+6}\) b/ \(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\) c/ \(\dfrac{x-1}{2x^2-2}-\dfrac{x+3}{4x+4}\)
d/ \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\) e/ \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
Bài 4 : Rút gọn và tính các giá trị của biểu thức
a/ \(\dfrac{3x^2-x}{9x^2-6x+1}\) tại x = \(\dfrac{1}{3}\) b/\(\dfrac{x^2-2xy+y^2-9}{x^2-xy+3x}\) Tại x = 2016 ; y = 3
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 \(x=-2\)
Bài 2:
\(a,\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
\(b,\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
\(c,\dfrac{x^2-9}{x^2+6x+9}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x-3}{x+3}\)
\(d,\dfrac{x^2+2x+1}{3x+3}=\dfrac{\left(x+1\right)^2}{3\left(x+1\right)}=\dfrac{x+1}{3}\)
\(e,\dfrac{x^2+7x+12}{x^2+5x+6}=\dfrac{x^2+2x+6x+12}{x^2+2x+3x+6}=\dfrac{x\left(x+2\right)+6\left(x+2\right)}{x\left(x+2\right)+3\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x+6\right)}{\left(x+2\right)\left(x+3\right)}=\dfrac{x+6}{x+3}\)
\(\left(a\right)\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
\(\left(b\right)\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
\(\left(c\right)\dfrac{x^2-9}{x^2+6x+9}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x-3}{x+3}\)
\(\left(d\right)\dfrac{x^2+2x+1}{3x+3}=\dfrac{\left(x+1\right)^2}{3\left(x+1\right)}=\dfrac{x+1}{3}\)
\(\left(e\right)\dfrac{x^2+7x+12}{x^2+5x+6}=\dfrac{x\left(x+3\right)+4\left(x+3\right)}{x\left(x+2\right)+3\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\)
\(a.\dfrac{15}{2x+6}+\dfrac{5x}{2x+6}=\dfrac{5\left(x+3\right)}{2\left(x+3\right)}=\dfrac{5}{2}\)
\(b.\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}=\dfrac{y^2}{xy\left(2x-y\right)}+\dfrac{-4x^2}{xy\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{\left(y-2x\right)\left(y+2x\right)}{-xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\)
\(c.\dfrac{x-1}{2x^2-2}-\dfrac{x+3}{4x+4}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{4\left(x+1\right)}=\dfrac{2x-2}{4\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{4\left(x-1\right)\left(x+1\right)}=\dfrac{2x-2-x^2-2x+3}{4\left(x-1\right)\left(x+1\right)}=\dfrac{1-x^2}{4\left(x-1\right)\left(x+1\right)}=\dfrac{-\left(x-1\right)\left(1+x\right)}{4\left(x-1\right)\left(1+x\right)}=\dfrac{-1}{4}\)
\(d.\dfrac{4y^2}{11x^4}.\left(\dfrac{-3x^2}{8y}\right)=\dfrac{y}{11x^2}.\left(\dfrac{-3}{2}\right)=\dfrac{-3y}{22x^2}\)
\(e.\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}=\dfrac{5\left(x+2\right)}{2\left(x-2\right)}.\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{5}{2}.\left(-2\right)=-5\)
\(a.\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
Thay \(x=\dfrac{1}{3}\Rightarrow\dfrac{\dfrac{1}{3}}{3.\dfrac{1}{3}-1}=0\)