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1) a) đkxđ \(x\ne\pm3,x\ne1\)
Ta có : \(P=\left(\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^2+3}{x^2-9}\right):\left(\frac{2x-2}{x-3}-1\right)\)
\(=\left(\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+3}{\left(x+3\right)\left(x-3\right)}\right):\frac{2x-2-x+3}{x-3}\)
\(=\frac{2x^2-6x+x^2+3x-3x^2-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+1}{x-3}\)
\(=\frac{-3x-3}{\left(x+3\right)\left(x-3\right)}.\frac{x-3}{x+1}=\frac{-3\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}=\frac{-3}{x+3}\)
b) Để \(P\in Z\) thì \(\frac{-3}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(-3\right)=\left\{\pm1,\pm3\right\}\)
Ta có bảng giá trị
| x+3 | 1 | -1 | 3 | -3 |
| x | -2 | -4 | 0 |
-6 |
Vậy với \(x\in\left\{-2,-4,0,6\right\}\) thì \(P\in Z\)
c) \(\left|x+3\right|=5\Leftrightarrow\left[{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Thay x=2 vào P, ta có : \(P=-\frac{3}{2+2}=-\frac{3}{4}\)
Thay x=-8 vào P, ta có : \(P=-\frac{3}{-8+2}=\frac{1}{2}\)
Vậy ....
2) a) đkxđ : \(x\ne1\)
Ta có : \(R=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=1:\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=1:\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+x+1}{x}\)
Xét : \(P-3=\frac{x^2+x+1}{x}-3=\frac{x^2-2x+1}{x}=\frac{\left(x-1\right)^2}{x}\)
+)Nếu \(x\ge0,x\ne1\Rightarrow R>3\)
+) Nếu \(x< 0\Rightarrow R< 3\)
+) Nếu \(\left[{}\begin{matrix}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{matrix}\right.\) \(\Rightarrow R=3\)
c) Để \(R>4\Rightarrow\frac{x^2+x+1}{x}>4\) \(\Rightarrow x^2+x+1>4x\)
\(\Rightarrow x^2>3x-1\) \(\Rightarrow x>\frac{3x-1}{x}=3-\frac{1}{x}\)
Vậy \(x>3-\frac{1}{x}thìR>4\)
d) Thay x=1/4 vào R, ta có : \(R=\frac{\frac{1}{16}+\frac{1}{4}+1}{\frac{1}{4}}=\frac{21}{4}\)
đề bài mk cảm thấy nó sao sao í bạn ạ hoặc do mk tính sai
a: \(A=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x^2-15x}-\dfrac{8}{5x^2+10x}\right):\dfrac{x^2-2x+2}{x^2-x-6}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x\left(x-3\right)}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4}{5x}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x+8-8}{5x\left(x+2\right)}\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x}{5x}\cdot\dfrac{x-3}{x^2-2x+2}=\dfrac{\left(5x+4\right)\left(x-3\right)}{5\left(x^2-2x+2\right)}\)
b: Khi x=1 thì \(A=\dfrac{\left(5+4\right)\left(1-3\right)}{5\left(1-2+2\right)}=\dfrac{9\cdot\left(-2\right)}{5}=\dfrac{-18}{5}\)
Khi x=3 thì \(A=\dfrac{\left(5\cdot3+4\right)\left(3-3\right)}{A}=0\)
Bài 3 : a, ĐK : \(x\ne0;-2\)
b, \(A=\frac{x^2}{x^2+2x}+\frac{2}{x+2}+\frac{2}{x}=\frac{x^2}{x\left(x+2\right)}+\frac{2x}{x\left(x+2\right)}+\frac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+4x+4}{x\left(x+2\right)}=\frac{\left(x+2\right)^2}{x\left(x+2\right)}=\frac{x+2}{x}\)
c, \(A=\frac{-\frac{3}{2}+2}{-\frac{3}{2}}=\frac{\frac{1}{2}}{-\frac{3}{2}}=-\frac{1}{12}\)
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
a: \(B=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x-3\right)\left(x-4\right)}=\dfrac{x-2}{x-4}\)
Để B>=0 thì x-4>0 hoặc x-2<=0
=>x>4 hoặc x<=2
b: Để B là số nguyên thì x-4+2 chia hết cho x-4
\(\Leftrightarrow x-4\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{5;6;2\right\}\)
1: ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
2: \(M=\left(\dfrac{x}{x-3}-\dfrac{x+3}{3x^2-6x-9}+\dfrac{1}{3x+3}\right)\cdot\dfrac{x^2-2x-3}{x^2+x+2}\)
\(=\left(\dfrac{x}{x-3}-\dfrac{x+3}{3\left(x-3\right)\left(x+1\right)}+\dfrac{1}{3\left(x+1\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+1\right)}{x^2+x+2}\)
\(=\dfrac{3x\left(x+1\right)-x-3+x-3}{3\left(x-3\right)\left(x+1\right)}\cdot\dfrac{\left(x-3\right)\left(x+1\right)}{x^2+x+2}\)
\(=\dfrac{3x^2+3x-6}{3}\cdot\dfrac{1}{x^2+x+2}=\dfrac{x^2+x-2}{x^2+x+2}\)
3: Vì x^2+x-2<x^2+x+2
nên M<1
a,\(M=\dfrac{x^2-6x+9}{x^2-7x+12}=\dfrac{\left(x-3\right)^2}{x^2-3x-4x+12}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-4\right)}\)
ĐKXĐ :\(\left\{{}\begin{matrix}x-3\ne0\\x-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne4\end{matrix}\right.\)
\(M=\dfrac{x-3}{x-4}\)
\(b,\left|x\right|=3\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Với x = 3 : \(M=\dfrac{3-3}{3-4}=\dfrac{0}{-1}=0\)
Với x = -3 : \(M=\dfrac{-3-3}{-3-4}=\dfrac{-6}{-7}=\dfrac{6}{7}\)
c, Thiếu đề bài
d, Để M = 5
\(\Rightarrow\dfrac{x-3}{x-4}=5\Rightarrow5x-20=x-3\)
\(\Rightarrow4x=17\Rightarrow x=\dfrac{17}{4}\)
Vậy....
e, \(M=\dfrac{x-3}{x-4}=\dfrac{x-4+1}{x-4}=1+\dfrac{1}{x-4}\)
Để M thuộc Z
\(\Rightarrow\dfrac{1}{x-4}\in Z\)
=> \(1⋮\left(x-4\right)\Rightarrow x-4\inƯ\left(1\right)=\left\{1;-1\right\}\)
Với x - 4 = 1 => x = 5 (t/m)
Với x - 4 = -1 => x = 3 (ko t/m)
Vậy x = 5 thì M thuộc Z
Cảm ơn bạn nha