Toán Lớp 8: Rút gọn biểu thức:
a) x+1/x-1 – x-1/x+1 – 4/1-x²
b) 4x/x+2 – 3x/x-2 + 12x/x²-4
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Toán Lớp 8: Rút gọn biểu thức:
a) x+1/x-1 – x-1/x+1 – 4/1-x²
b) 4x/x+2 – 3x/x-2 + 12x/x²-4
Comments ( 2 )
Giải đáp:
a) 4/{x – 1}
b) x/{x + 2}
Lời giải và giải thích chi tiết:
a)
{x + 1}/{x – 1} – {x – 1}/{x + 1} – 4/{1 – x^2} (ĐKXĐ: x ne +- 1)
= {x + 1}/{x – 1} – {x – 1}/{x + 1} – 4/{- (x^2 – 1)}
= {x + 1}/{x – 1} – {x – 1}/{x + 1} + 4/{x^2 – 1}
= {x + 1}/{x – 1} – {x – 1}/{x + 1} + 4/{(x – 1)(x + 1)}
= {(x + 1)(x + 1)}/{(x – 1)(x + 1)} – {(x – 1)(x – 1)}/{(x – 1)(x + 1)} + 4/{(x – 1)(x + 1)}
= {(x + 1)^2}/{(x – 1)(x + 1)} – {(x – 1)^2}/{(x – 1)(x + 1)} + 4/{(x – 1)(x + 1)}
= {x^2 + 2x + 1}/{(x – 1)(x + 1)} – {x^2 – 2x + 1}/{(x – 1)(x + 1)} + 4/{(x – 1)(x + 1)}
= {x^2 + 2x + 1 – (x^2 – 2x + 1) + 4}/{(x – 1)(x + 1)}
= {x^2 + 2x + 1 – x^2 + 2x – 1 + 4}/{(x – 1)(x + 1)}
= {(x^2 – x^2) + (2x + 2x) + (1 – 1 + 4)}/{(x – 1)(x + 1)}
= {4x + 4}/{(x – 1)(x + 1)}
= {4(x + 1)}/{(x – 1)(x + 1)}
= 4/{x – 1}
$\\$
b)
{4x}/{x + 2} – {3x}/{x – 2} + {12x}/{x^2 – 4} (ĐKXĐ: x ne +- 2)
= {4x}/{x + 2} – {3x}/{x – 2} + {12x}/{x^2 – 2^2}
= {4x}/{x + 2} – {3x}/{x – 2} + {12x}/{(x – 2)(x + 2)}
= {4x(x – 2)}/{(x – 2)(x + 2)} – {3x(x + 2)}/{(x – 2)(x + 2)} + {12x}/{(x – 2)(x + 2)}
= {4x^2 – 8x}/{(x – 2)(x + 2)} – {3x^2 + 6x}/{(x – 2)(x + 2)} + {12x}/{(x – 2)(x + 2)}
= {4x^2 – 8x – (3x^2 + 6x) + 12x}/{(x – 2)(x + 2)}
= {4x^2 – 8x – 3x^2 – 6x + 12x}/{(x – 2)(x + 2)}
= {(4x^2 – 3x^2) + (12x – 8x – 6x)}/{(x – 2)(x + 2)}
= {x^2 – 2x}/{(x – 2)(x + 2)}
= {x(x – 2)}/{(x – 2)(x + 2)}
= x/{x + 2}
#SunHee
$a)$ $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{1-x^2}$ \text{;(x ≠ ±1)}
$=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{-4}{\left(x-1\right)\left(x+1\right)}$
$=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-\left(-4\right)}{\left(x-1\right)\left(x+1\right)}$
$=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+4}{\left(x-1\right)\left(x+1\right)}$
$=\dfrac{4x+4}{\left(x-1\right)\left(x+1\right)}$
$=\dfrac{4}{x-1}$
$b)$ $\dfrac{4x}{x+2}-\dfrac{3x}{x-2}+\dfrac{12x}{x^2-4}$ \text{;(x ≠ ±2)}
$=\dfrac{4x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{3x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{12x}{\left(x+2\right)\left(x-2\right)}$
$=\dfrac{4x\left(x-2\right)-3x\left(x+2\right)+12x}{\left(x+2\right)\left(x-2\right)}$
$=\dfrac{x^2-2x}{\left(x+2\right)\left(x-2\right)}$
$=\dfrac{x}{x+2}$