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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 \(x=-2\)
\(a.\)
\(C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\) \(\left(1\right)\)
\(\text{Đ}KX\text{Đ}:\) \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\) \(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x-1}{x-3}\)
\(b\)
\(C=\dfrac{x-1}{x-3}=\dfrac{\left(x-3\right)+4}{x-3}=1+\dfrac{4}{x-3}\)
Để C nguyên thì \(x-3\in\text{Ư}\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-1;1;2;4;5;7\right\}\)
\(a.C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\) ( x # 2 ; x # 0 ; x # 3 )
\(C=\dfrac{2x^2-9x}{x\left(x-2\right)\left(x-3\right)}-\dfrac{x\left(x^2-9\right)}{x\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x^2-2x\right)\left(2x+1\right)}{x\left(x-2\right)\left(x-3\right)}\) \(C=\dfrac{2x^2-9x-x^3+9x+2x^3-3x^2-2x}{x\left(x-2\right)\left(x-3\right)}\)
\(C=\dfrac{x^3-x^2-2x}{x\left(x-2\right)\left(x-3\right)}\)
\(C=\dfrac{x\left(x-2\right)\left(x+1\right)}{x\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b. \(C=\dfrac{x+1}{x-3}=\dfrac{x-3+4}{x-3}=1+\dfrac{4}{x-3}\)
Để : C ∈ Z ⇒ ( x - 3 )∈ { 1 ; -1 ; 2 ; -2 ; 4 ; -4 } x - 3 x 1 4 - 1 2 (TM) 2 5(TM) -2 1(TM) 4 7(TM) -4 -1(TM) (KTM)
Vậy ,....
a: \(B=\left(\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{x-1}\right)\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3x-3-2}{x-1}\)
\(=\dfrac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-5}=\dfrac{-1}{2x-3}\)
b: Để B>0 thì 2x-3<0
hay x<3/2
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(C=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Để C là số nguyên thì \(x-3+4⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{4;5;1;7;-1\right\}\)
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b)
\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)
c)
Để \(P\le1\) thì:
\(-\dfrac{x}{x-3}\le1\)
\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\le1\)
`HaNa♬D`
Làm lại nha cái này đúng, kia sai nha=)
b)
Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)
\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)
c)
Để P\(\ge1\) thì:
\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\ge1\)
`HaNa☘D`