第25页 - 通城学典课时作业本七年级数学苏科版江苏专用

$9m^2+6mn^2+n^4$

1

$ \begin{aligned}解：原式&=(50-0.2)^2 \\ &=50^2-2×50×0.2+0.2^2 \\ &=2500-20+0.04 \\ &=2480.04 \\ \end{aligned}$

$ \begin{aligned}解：原式&=(100+11)^2-10000-121 \\ &=100^2+2×100×11+11^2-100^2-11^2 \\ &=2200 \\ \end{aligned}$

$ \begin{aligned}解：原式&=-(2a+7b)^2 \\ &=-(4a^2+28ab+49b^2) \\ &=-4a^2-28ab-49b^2 \\ \end{aligned}$

$ \begin{aligned} 解：原式&=(x-2y)^2+2z(x-2y)+z^2 \\ &=x^2-4xy+4y^2+2xz-4yz+z^2 \\ \end{aligned}$

$解：因为(3x+y)^2=25，(3x−y)^2=9，$

$所以9x^2+6xy+y^2=25，9x^2−6xy+y²=9.$

$将上述两个等式相减，得6xy-(-6xy)=25-9，$

$即12xy=16，$

$所以xy=\frac{4}{3}$

$(2×5+1)^2=(6×10+1)^2−(6×10)^2$

$解：(2)第n个等式：(2n+1)^2=[2n(n+1)+1]^2-[2n(n+1)]^2$

$理由：左边=4n^2+4n+1$

$右边=[2n(n+1)]^2+2×2n(n+1)×1+1^2-[2n(n+1)]^2$

$=4n(n+1)+1$

$=4n^2+4n+1$

$∴左边=右边$

$∴等式成立$