Trừ hai phân thức :
\(\frac{x+3}{x^2-1}\)- \(\frac{x+1}{x^2-x}\)
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ĐK: \(x\ne1\)
\(\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-2x}{x^2+x+1}-\frac{6}{x-1}\)
\(=\frac{4x^2-3x+5-\left(1-2x\right)\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+5+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)
\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)
\(=\frac{3x-2}{9x^2-4}\)
\(=\frac{1}{3x+2}\)
\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)
\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)
\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)
\(=\frac{\left(-x^2-3\right)\left(x-3\right)}{\left(x-3^2\right)\left(x+3\right)}\)
\(=\frac{-x^2-3}{\left(x-3\right)\left(x+3\right)}\)
học tốt
\(\frac{1}{x-3}-\frac{3}{2x+6}-\frac{x}{2x^2-12x+18}\)
\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x^2-6x+9\right)}\)
\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x-3\right)^2}\)
\(=\frac{2\left(x-3\right)\left(x+3\right)-3\left(x-3\right)^2-x\left(x+3\right)}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{2\left(x^2-9\right)-3\left(x^2-6x+9\right)-x\left(x+3\right)}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{2x^2-18-3x^2+18x-27-x^2-3x}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{-2x^2+15x-45}{2\left(x-3\right)^2\left(x+3\right)}\)
Ta có:
\(\dfrac{x^2-4}{x+1}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{x+1}\)
Và:
\(\dfrac{x+2}{2x}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{2x\left(x-2\right)}\)
Vậy ta đã biến đổi hai phân thức đó để chúng bằng phân thức cũ và có tủ bằng nhau
ĐKXĐ : \(x\ne-1\)
\(\frac{1}{2x+2}-\frac{x-1}{3x^2+6x+3}\)
\(=\frac{1}{2\left(x+1\right)}-\frac{x-1}{3\left(x^2+2x+1\right)}\)
\(=\frac{1}{2\left(x+1\right)}-\frac{x-1}{3\left(x+1\right)^2}\)
\(=\frac{3\left(x+1\right)}{2\left(x+1\right)\cdot3\left(x+1\right)}-\frac{2\left(x-1\right)}{3\left(x+1\right)^2\cdot2}\)
\(=\frac{3x+3}{6\left(x+1\right)^2}-\frac{2x-2}{6\left(x+1\right)^2}\)
\(=\frac{3x+3-2x+2}{6\left(x+1\right)^2}\)
\(=\frac{x+5}{6\left(x+1\right)^2}\)
1) Biến đổi A, ta được:
\(A=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)
Do đó:
\(A< 1\Rightarrow1+\frac{7}{x-2}< 1\Rightarrow\frac{7}{x-2}< 0\left(1\right)\)
Mà 7>0 nên:
\(\left(1\right)\Rightarrow x-2< 0\Rightarrow x< 2\)
2)
+) Biến đổi B, ta được:
\(B=\frac{3\left(x-2\right)+2x^2-x-19-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\frac{3x-6+2x^2-x-19-x^2-2x}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-25}{x^2-4}\left(đpcm\right)\)
+) Từ 1) và 2), ta suy ra:
\(P=\frac{B}{A}=\frac{\frac{x+5}{x-2}}{\frac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)\left(x+2\right)}}=\frac{1}{\frac{x-5}{x+2}}=\frac{x+2}{x-5}\)
3) Biến đổi P, ta được:
\(P=\frac{x-5+3}{x-5}=1+\frac{3}{x-5}\)
P nguyên khi và chỉ khi \(\frac{3}{x-5}\) nguyên, hay \(x-5\inƯ\left(3\right)\)
Ta có bảng:
x-5 | -3 | -1 | 1 | 3 |
x | 2 | 4 | 6 | 8 |
Vậy ta có 4 giá trị của x trên thoả mãn đề bài.
Chúc bạn học tốt nha
ĐK : \(x\ne0;\pm1\)
\(\frac{x+3}{x^2-1}-\frac{x+1}{x^2-x}=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{x\left(x-1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+3x-x^2+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3x+1}{x\left(x-1\right)\left(x+1\right)}\)