[Toán 8] Rút gọn phân thức

H

hothinhuthao

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I

iceghost

Câu $A$ có sai đề không bạn
$B=\dfrac{ a^3 - 3a + 2 }{ 2a^3 - 7a^2 + 8a -3 } \\
=\dfrac{a^3-a-2a+2}{2a^3-3a^2-4a^2+6a+2a-3} \\
=\dfrac{a(a^2-1)-2(a-1)}{a^2(2a-3)-2a(2a-3)+(2a-3)} \\
=\dfrac{a(a-1)(a+1)-2(a-1)}{(a^2-2a+1)(2a-3)} \\
=\dfrac{(a-1)(a^2+a)-2(a-1)}{(a^2-2a+1)(2a-3)} \\
=\dfrac{(a-1)(a^2+a-2)}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)(a^2+2a-a-2)}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)[a(a+2)-(a+2)]}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)(a-1)(a+2)}{(a-1)(a-1)(2a-3)} \\
=\dfrac{a+2}{2a-3}$
 
I

iceghost

Câu $A$ có sai đề không bạn
$B=\dfrac{ a^3 - 3a + 2 }{ 2a^3 - 7a^2 + 8a -3 } \\
=\dfrac{a^3-a-2a+2}{2a^3-3a^2-4a^2+6a+2a-3} \\
=\dfrac{a(a^2-1)-2(a-1)}{a^2(2a-3)-2a(2a-3)+(2a-3)} \\
=\dfrac{a(a-1)(a+1)-2(a-1)}{(a^2-2a+1)(2a-3)} \\
=\dfrac{(a-1)(a^2+a)-2(a-1)}{(a^2-2a+1)(2a-3)} \\
=\dfrac{(a-1)(a^2+a-2)}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)(a^2+2a-a-2)}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)[a(a+2)-(a+2)]}{(a-1)^2(2a-3)} \\
=\dfrac{(a-1)(a-1)(a+2)}{(a-1)(a-1)(2a-3)} \\
=\dfrac{a+2}{2a-3}$
Mình nghĩ đề đúng chắc là :
$A=\dfrac{ x^4 + x^3 + x +1 }{ x^4 - x^3 + 2x^2 - x+1 } \\
=\dfrac{x^3(x+1)+(x+1)}{x^4+x^2-x^3-x+x^2+1} \\
=\dfrac{(x+1)(x^3+1)}{x^2(x^2+1)-x(x^2+1)+(x^2+1)} \\
=\dfrac{(x+1)(x+1)(x^2-x+1)}{(x^2+1)(x^2-x+1)} \\
=\dfrac{(x+1)^2}{x^2+1}$
 
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