第95页 - 通城学典课时作业本八年级数学苏科版江苏专版

解：原式=\frac {4m+5+(m-1)(m+1)}{m+1}÷\frac {m+2}{m+1}

=\frac {4m+5+\ \mathrm {m^2}-1}{m+1}×\frac {m+1}{m+2}

=\frac {(m+2)^2}{m+1}×\frac {m+1}{m+2}

=m+2

解：原式=\frac {a+b-b}{(a+b)(a-b)}×\frac {(a-b)^2}{a(a-b)}

=\frac 1{a+b}

$ 解：原式= \frac {a-4}{a} ÷ [\frac {a+2}{a(a-2)} - \frac {a-1}{(a-2)²}] $

$ = \frac {a-4}{a} ÷[ \frac {(a+2)(a-2)}{a(a-2)²} -\frac {a(a-1)}{a(a-2)²} ]$

$ = \frac {a-4}{a} ÷ \frac {a²-4-a²+a}{a(a-2)²}$

$= \frac {a-4}{a} \cdot \frac {a(a-2)²}{a-4} $

$ =(a-2)²$

$ =a²-4a+4$

$ ∵a²-( \frac {1}{4} ) ^{-1} \cdot a+3=0$

$ ∴a²-4a+3=0$

$ ∴a²-4a=-3$

$ ∴原式=-3+4=1$

$解：∵\frac {2}{1×3}=\frac 11-\frac 13，\frac 2{2×4}=\frac 12-\frac 14，\frac 2{3×5}=\frac 13-\frac 15···$

$∴\frac 2{n(n+2)}=\frac 1n-\frac 1{n+2}$

$∴原式=\frac 12(1-\frac 13+\frac 12-\frac 14+\frac 13-\frac 15+···+\frac 1n-\frac 1{n+2})$

$ =\frac 12×(1+\frac 12-\frac1{n+1}-\frac 1{n+2} )$

$ =\frac {3n^2+5n}{4(n+1)(n+2)}$

D

$ -\frac 12$

$ \begin{aligned} 解：3-x-1&=x-4 \\ x&=3 \\ \end{aligned}$

$ 检验：当x=3时，x-4≠0$

$ ∴x=3是方程的解$

$ \begin{aligned}解：x(x+2)-(x-1)(x+2)&=3 \\ x^2+2x-x^2-2x+x+2&=3 \\ x&=1 \\ \end{aligned}$

$检验：当x=1时，(x-1)(x+2)=0$

$∴x=1是增根，原方程无解$