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a) `(x^3-x^2)/(x^3-2x^2+x)`
`=(x^2(x-1))/(x(x-1)(x-1))`
`=x/(x-1)`
`=>` 2 phân thức bằng nhau.
b) `(x^2+2x+1)/(2x^2-2)`
`=((x+1)(x+1))/(2(x+1)(x-1))`
`=(x+1)/(2(x-1))`
`=(x+1)/(2x-2)`
`=>` 2 phân thức bằng nhau
a) Ta có: \(\dfrac{x^3-x^2}{x^3-2x^2+x}\)
\(=\dfrac{x^2\left(x-1\right)}{x\left(x^2-2x+1\right)}\)
\(=\dfrac{x\cdot\left(x-1\right)}{\left(x-1\right)^2}=\dfrac{x}{x-1}\)
b) Ta có: \(\dfrac{x^2+2x+1}{2x^2-2}\)
\(=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{2x-2}\)
\(\left\{{}\begin{matrix}\dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\\dfrac{x-3}{x}\\\dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\end{matrix}\right.\)
Vậy \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)
\(ĐK:x\ne0;x\ne\pm1\\ \dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\ \dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\)
Do đó 3 phân thức trên bằng nhau
\(\text{Ta có : }\dfrac{x^2-2x-3}{x^2+x}\\ =\dfrac{x^2+x-3x-3}{x\left(x+1\right)}\\ =\dfrac{\left(x^2+x\right)-\left(3x+3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)}\\ \\ =\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\text{ }\left(1\right)\)
\(\dfrac{x^2-4x+3}{x^2-x}\\ =\dfrac{x^2-x-3x+3}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x^2-x\right)-\left(3x-3\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)
Vậy 3 phân thức \(\dfrac{x^2-2x-3}{x^2+x};\dfrac{x-3}{x};\dfrac{x^2-4x+3}{x^2-x}\) bằng nhau
Giả sử :
\(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)
\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{x-3}{x}=\dfrac{x-3}{x}=\dfrac{x-3}{x}\)
Vậy 3 thức trên bằng nhau
Bài 2:
(1 + x)3 + (1 - x)3 - 6x(x + 1) = 6
<=> x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 - 6x = 6
<=> -6x + 2 = 6
<=> -6x = 6 - 2
<=> -6x = 4
<=> x = -4/6 = -2/3
Bài 3:
a) (7x - 2x)(2x - 1)(x + 3) = 0
<=> 10x3 + 25x2 - 15x = 0
<=> 5x(2x - 1)(x + 3) = 0
<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0
<=> x = 0 hoặc x = 1/2 hoặc x = -3
b) (4x - 1)(x - 3) - (x - 3)(5x + 2) = 0
<=> 4x2 - 13x + 3 - 5x2 + 13x + 6 = 0
<=> -x2 + 9 = 0
<=> -x2 = -9
<=> x2 = 9
<=> x = +-3
c) (x + 4)(5x + 9) - x2 + 16 = 0
<=> 5x2 + 9x + 20x + 36 - x2 + 16 = 0
<=> 4x2 + 29x + 52 = 0
<=> 4x2 + 13x + 16x + 52 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> 4x + 13 = 0 hoặc x + 4 = 0
<=> x = -13/4 hoặc x = -4