Toán Lớp 8: Rút gọn đa thức sau:
$\frac{x+1}{x-2}$ + $\frac{4}{x+2}$ + $\frac{2-7x}{x^2-4}$
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Toán Lớp 8: Rút gọn đa thức sau:
$\frac{x+1}{x-2}$ + $\frac{4}{x+2}$ + $\frac{2-7x}{x^2-4}$
Thanks
Comments ( 2 )
$\text{Bài làm:}$
$\text{ $\frac{x + 1}{x – 2}$ $+$$\frac{4}{x + 2}$ $+$$\frac{2 – 7x}{x^{2} – 4}$ }$
$\text{ = $\frac {x + 1}{x – 2}$ $+$$\frac{4}{x + 2}$ $+$$\frac{2 – 7x}{(x – 2 ). ( x + 2)}$}$
$\text{ = $\frac{( x + 2) . ( x + 1 ) + 4(x – 2) + 2 – 7x}{( x – 2). (x +2)}$ }$
$\text{=$\frac{x^{2} + x + 2x + 2 + 4( x – 2 ) + 2 – 7x}{(x – 2). ( x + 2)}$ }$
$\text{=$\frac{x^{2} + x + 2x + 2 + 4x – 8 + 2 – 7x}{( x – 2). ( x +2)}$ }$
$\text{=$\frac{x^{2}+ 0 +2 – 8 + 2 }{( x – 2). ( x + 2)}$ }$
$\text{=$\frac{x^{2} + 0 – 4}{( x – 2 ). ( x + 2)}$ }$
$\text{=$\frac{x^{2} – 4}{( x – 2 ). ( x + 2)}$}$
$\text{= $\frac{( x – 2). ( x + 2)}{( x – 2). ( x – 2)}$ }$
$\text{= 1}$
$#thuanhuy$
$\text{Gửi bn!}$
{x + 1}/{x – 2} + {4}/{x + 2} + {2 – 7x}/{x^2 – 4} $[MTC: (x – 2)(x + 2)]$
= {x + 1}/{x – 2} + {4}/{x + 2} + {2 – 7x}/{x^2 – 2^2}
= {x + 1}/{x – 2} + {4}/{x + 2} + {2 – 7x}/{(x – 2)(x + 2)}
= {(x + 1).(x + 2) + 4.(x – 2) + 2 – 7x}/{(x – 2)(x + 2)}
= {x^2 + 2x + x + 2 + 4x – 8 + 2 – 7x}/{(x – 2)(x + 2)}
= {x^2 + 2x + x + 4x – 7x + 2 – 8 + 2}/{(x – 2)(x + 2)}
= {x^2 – 4}/{(x – 2)(x + 2)}
= {x^2 – 2^2}/{(x – 2)(x + 2)}
= {(x – 2)(x + 2)}/{(x – 2)(x + 2)}
= 1