第43页 - 启东中学作业本七年级数学苏科版宿迁专版

$ \begin{aligned}解：原式&=t^{2m+4}-4t^{2m+4} \\ &=-3t^{2m+4}. \\ \end{aligned}$

$ \begin{aligned}解：原式&=-b^4-b^4 \\ &=-2b^4. \\ \end{aligned}$

$ \begin{aligned}解：原式&=9÷8-\frac{1}{4} \\ &=\frac{7}{8} \\ \end{aligned}$

$ \begin{aligned}解：原式&=\frac{1}{2}+\frac{1}{4}-1 \\ &=-\frac{1}{4} \\ \end{aligned}$

$ \begin{aligned}解：原式&=(0.25×4)^9×(2×5)^{20} \\ &=10^{20}. \\ \end{aligned}$

$ \begin{aligned}解：原式&=2^n×1010^n×(\frac{1}{1010})^{n+1}×(\frac{1}{2})^{n+2} \\ &=\frac{1}{4}× \frac{1}{1010} \\ &=\frac{1}{4040} \\ \end{aligned}$

$解：原方程可变形为3×5^{x+1}=75，$

$即5^{x+1}=5^2，$

$ x+1=2，解得x=1.$

$解：原方程可变形为2^2×2^{2x+1}-2^{2x+1}=3×2^6，$

$即3×2^{2x+1}=3×2^6，2x+1=6，$

$解得x=\frac{5}{2}.$

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A

$ \begin{aligned}解：(1)原式&=(10^a)^2+(10^b)^3 \\ &=4^2+3^3 \\ &=16+27 \\ &=43. \\ \end{aligned}$

$ \begin{aligned}(2)原式&=(10^a)^2×(10^b)^3÷10 \\ &=4^2×3^3÷10 \\ &=43.2. \\ \end{aligned}$