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a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)
Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)
\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)
\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)
\(=\dfrac{3x}{x+3}\)
b) Ta có: \(x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được:
\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)
Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)
a + b , ĐKXĐ : \(x\ne2;-3\)
\(A=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-3}{x-2}\)
c, Thay x = 2 ta có : ... Vì ko thỏa mãn giá trị của phân thức x khác 2 nên ko có giá trị biểu thức
d, Ta có : \(\frac{x-3}{x-2}=\frac{x-2-1}{x-2}=-\frac{1}{x-2}\)
\(-x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
| -x + 2 | 1 | -1 |
| x | 1 | 3 |
a: \(P=\dfrac{x^2+6x+9-x^2+6x-9-4}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-1}{x-3}\)
\(=\dfrac{4\left(3x-1\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{3x-1}=\dfrac{4}{x+3}\)
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
a) Điều kiện: \(x\ne\pm1\)
\(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)
\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)
b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Khi \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)
Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)
c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)
