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\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
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)B = \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)
= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)
= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)
b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)
Thay x = -4 vào B, ta có:
B = \(\dfrac{-4.3}{-4+3}=12\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)
<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)
d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên
<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)
| x+3 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -12(C) | -6(C) | -4(C) | -2(C) | 0(C) | 6(C) |
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: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
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: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
\(A=\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\left(ĐKXĐ:x\ne\pm3\right)\)
a, \(A=\dfrac{-\left(x-3\right)\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)}+\dfrac{x}{x+3}\)
\(=-1+\dfrac{x}{x+3}=\dfrac{-x-3+x}{x+3}=\dfrac{-3}{x+3}\)
b, \(x^2-2x-3=0\Leftrightarrow x^2-3x+x-3\Leftrightarrow x\left(x-3\right)+\left(x-3\right)\Leftrightarrow\left(x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
TH1 : Nếu x = 3 thì gt của biểu thức \(A=\dfrac{-3}{3+3}=-\dfrac{3}{6}=-\dfrac{1}{2}\)
TH2 : Nếu x = -2 thì gt của biểu thức \(A=\dfrac{-3}{-2+3}=-3\)
c, Để A nhận giá trị nguyên thì \(x+3\inƯ\left(3\right)\) ( Ư(-3 ) cũng được như nhau nhé ! )
Xét bảng :
| x + 3 | x |
| 1 | -2 |
| -1 | -4 |
| 3 | 0 |
| -3 | -6 |
Vậy để A nguyên thì \(x\in\left\{-6;-4;-2;0\right\}\)
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)
\(a.\)
\(C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\) \(\left(1\right)\)
\(\text{Đ}KX\text{Đ}:\) \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\) \(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x-1}{x-3}\)
\(b\)
\(C=\dfrac{x-1}{x-3}=\dfrac{\left(x-3\right)+4}{x-3}=1+\dfrac{4}{x-3}\)
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 }
Vậy ,....