Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{5x+10+14x-28-20}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19}{2\left(x+2\right)}\\ c,x=-\dfrac{1}{2}\Leftrightarrow A=\dfrac{19}{2\left(2-\dfrac{1}{2}\right)}=\dfrac{19}{2\cdot\dfrac{3}{2}}=\dfrac{19}{3}\)
a: ĐKXĐ: x^3-3x-2<>0
=>x^3-x-2x-2<>0
=>x(x-1)(x+1)-2(x+1)<>0
=>(x+1)(x-2)(x+1)<>0
=>x<>2 và x<>-1
b: \(A=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x-2\right)\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x-2}\)
c:
A<1
=>A-1<0
\(A-1=\dfrac{x^2-2x+1-x+2}{x-2}=\dfrac{x^2-3x+3}{x-2}\)
=>x-2<0
=>x<2
a: DKXĐ: x^3-3x-2<>0
=>x^3-x-2x-2<>0
=>x(x-1)(x+1)-2(x+1)<>0
=>(x+1)(x^2-x-2)<>0
=>(x+1)(x-2)(x+1)<>0
=>\(x\notin\left\{2;-1\right\}\)
b: \(A=\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\dfrac{\left(x-1\right)^2}{x-2}\)
c: Để A<1 thì A-1<0
=>\(\dfrac{x^2-2x+1-x+2}{x-2}< 0\)
=>x-2<0
=>x<2
a) ĐK: \(x\ne4,x\ne2;x\ne-2\)
b) \(A=\dfrac{x^3}{x-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(A=\dfrac{x^3}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^3-x^2-2x-2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^3-x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{\left(x-1\right)\left(x^2-4\right)}{x^2-4}\)
\(A=x-1\)
c) \(A=0\) khi:
\(x-1=0\)
\(\Leftrightarrow x=1\left(tm\right)\)
d) A dương khi: \(A>0\)
\(x-1>0\)
\(\Leftrightarrow x>1\)
Kết hợp với đk:
\(x>1,x\ne4,x\ne2\)
a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)
Ta có : \(P=\dfrac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{4}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x\left(x-3\right)+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x-x+6}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)
b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)
- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)
Vậy ...
a ĐKXĐ : \(x\ne2,x\ne3\)
\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)
Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)
\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\)
Thử lại ta thấy đúng
Vậy...
a) P có nghĩa khi \(\hept{\begin{matrix}2x+4\ne0\\2x-4\ne0\\x^2-4\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow\hept{\begin{matrix}2\left(x+2\right)\ne0\\2\left(x-2\right)\ne0\\\left(x-2\right)\left(x+2\right)\ne0\\x-2\ne0\end{matrix}\Leftrightarrow\hept{\begin{matrix}x+2\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow x\ne\pm2}\)
vậy P có nghĩa khi \(x\ne\pm2\)
b) \(P=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}-\frac{8}{x^2-4}\right):\frac{4}{x-2}\left(x\ne\pm2\right)\)
\(\Leftrightarrow P=\left(\frac{x+2}{2\left(x-2\right)}+\frac{x-2}{2\left(x+2\right)}-\frac{8}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\left[\frac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\left[\frac{x^2+4x+4}{2\left(x-2\right)\left(x+2\right)}+\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\frac{2x^2-8}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}=\frac{2\left(x^2-4\right)\left(x-2\right)}{8\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(x-2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{x-2}{4}\)
vậy P=\(\frac{x-2}{4}\left(x\ne\pm2\right)\)