Toán Lớp 8: Làm tính cộng: $a)$ $\frac{2}{x^{2}+2x}+$ $\frac{3x^{2}-6x}{x^{2}-2x+4}+$$\frac{10x^{2}+28x-8}{x^{4}+8x}$
$b)$ $\frac{1}{2x-2}+$ $\frac{x+1}{x^{2}+x+1}+$ $\frac{1-3x-x^{2}}{2x^{3}-2}$
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Toán Lớp 8: Làm tính cộng: $a)$ $\frac{2}{x^{2}+2x}+$ $\frac{3x^{2}-6x}{x^{2}-2x+4}+$$\frac{10x^{2}+28x-8}{x^{4}+8x}$
$b)$ $\frac{1}{2x-2}+$ $\frac{x+1}{x^{2}+x+1}+$ $\frac{1-3x-x^{2}}{2x^{3}-2}$
Comments ( 2 )
Giải đáp+giải thích các bước giải:
a,
\frac{2}{x^2+2x}+\frac{3x^2-6x}{x^2-2x+4}+\frac{10x^2+28x-8}{x^4+8x} (ĐKXĐ:x\ne0;x\ne-2)
=\frac{2}{x(x+2)}+\frac{3x(x-2)}{x^2-2x+4}+\frac{10x^2+28x-8}{x(x^3+8)}
=\frac{2}{x(x+2)}+\frac{3x(x-2)}{x^2-2x+4}+\frac{10x^2+28x-8}{x(x+2)(x^2-2x+4)}
=\frac{2(x^2-2x+4)}{x(x+2)(x^2-2x+4)}+\frac{3x^2(x-2)(x+2)}{x(x+2)(x^2-2x+4)}
+\frac{10x^2+28x-8}{x(x+2)(x^2-2x+4)}
=\frac{2x^2-4x+8+3x^2(x^2-4)+10x^2+28x-8}{x(x+2)(x^2-2x+4)}
=\frac{2x^2-4x+8+3x^4-12x^2+10x^2+28x-8}{x(x+2)(x^2-2x+4)}
=\frac{3x^4+24x}{x(x+2)(x^2-2x+4)}
=\frac{3x(x^3+8)}{x(x+2)(x^2-2x+4)}
=\frac{3x(x+2)(x^2-2x+4)}{x(x+2)(x^2-2x+4)}
=3
b,
\frac{1}{2x-2}+\frac{x+1}{x^2+x+1}+\frac{1-3x-x^2}{2x^3-2} (ĐKXĐ: x\ne1)
=\frac{1}{2(x-1)}+\frac{x+1}{x^2+x+1}+\frac{1-3x-x^2}{2(x^3-1)}
=\frac{x^2+x+1}{2(x-1)(x^2+x+1)}+\frac{2(x+1)(x-1)}{2(x-1)(x^2+x+1)}
+\frac{1-3x-x^2}{2(x-1)(x^2+x+1)}
=\frac{x^2+x+1+2(x^2-1)+1-3x-x^2}{2(x-1)(x^2+x+1)}
=\frac{x^2+x+1+2x^2-2+1-3x-x^2}{2(x-1)(x^2+x+1)}
=\frac{2x^2-2x}{2(x-1)(x^2+x+1)}
=\frac{2x(x-1)}{2(x-1)(x^2+x+1)}
=\frac{x}{x^2+x+1}
Giải đáp:
a)
ĐKXĐ : x \ne 0; x \ne -2.
2/(x^(2)+2x)+(3x^(2)-6x)/(x^(2)-2x+4)+(10x^(2)+28x-8)/(x^(4)+8x)
=2/[x(x+2)]+(3x^(2)-6x)/(x^(2)-2x+4)+(10x^(2)+28x-8)/[x(x^(3)+8)]
=[2(x^(2)-2x+4)]/[x(x+2)(x^(2)-2x+4)]+[x(x+2)(3x^(2)-6x)]/[x(x+2)(x^(2)-2x+4)]+(10x^(2)+28x-8)/[x(x+2)(x^(2)-2x+4)]
=(2x^(2)-4x+8+3x^(2)(x+2)(x-2)+10x^(2)+28x-8)/[x(x+2)(x^(2)-2x+4)]
=(2x^(2)-4x+8+3x^(4)-12x^(2)+10x^(2)+28x-8)/[x(x+2)(x^(2)-2x+4)]
=(3x^(4)+24x)/[x(x+2)(x^(2)-2x+4)]
=(3x(x^(3)+8))/[x(x^(3)+8)]
=3
b)
ĐKXĐ : x \ne 1.
1/(2x-2)+(x+1)/(x^(2)+x+1)+(1-3x-x^2)/(2x^(3)-2)
=1/[2(x-1)]+(x+1)/(x^(2)+x+1)+(1-3x-x^2)/[2(x^(3)-1)]
=(x^(2)+x+1)/[2(x-1)(x^(2)+x+1)]+[2(x+1)(x-1)]/[2(x-1)(x^(2)+x+1)]+(1-3x-x^2)/[2(x-1)(x^(2)+x+1)]
=(x^(2)+x+1+2x^(2)-2+1-3x-x^2)/[2(x-1)(x^(2)+x+1)]
=(2x^(2)-2x)/[2(x-1)(x^(2)+x+1)]
=[2x(x-1)]/[2(x-1)(x^(2)+x+1)]
=x/(x^(2)+x+1)