Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)
ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)
b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)
\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)
a: Thay x=5 vào B, ta được:
\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)
b: \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
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: \(C=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
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: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
\(a.\)
\(C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\) \(\left(1\right)\)
\(\text{Đ}KX\text{Đ}:\) \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\) \(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x-1}{x-3}\)
\(b\)
\(C=\dfrac{x-1}{x-3}=\dfrac{\left(x-3\right)+4}{x-3}=1+\dfrac{4}{x-3}\)
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 ,....