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a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5
a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)=\left(\dfrac{1-2x+x^2}{x\left(x+1\right)}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\left(\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\right)\cdot\left(\dfrac{x}{\left(x-1\right)^2}\right)=\dfrac{\left(x-1\right)^2\cdot x}{\left(x-1\right)^2\cdot x\cdot\left(x+1\right)}=\dfrac{1}{x+1}\)
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
