第24页 - 苏科版七年级伴你学数学答案（上下册）

${n}^{3}-4n$

$12{m}^{2}+11mn+2{n}^{2}$

$解：原式=6m^2+15m-2m-5$

$ =6m^2+13m-5$

$解：原式=6{x}^{2}-15xy-2xy+5{y}^{2}$

$=6{x}^{2}-17xy+5{y}^{2}$

$解：原式={m}^{3}-2{m}^{2}+4m+2{m}^{2}-4m+8$

$ ={m}^{3}+8$

解：原式$=3({x}^2+2x-2x-4)-(3{x}^2-3x+4x-4)$

$=3({x}^2-4)-(3{x}^2+x-4)$

$=3{x}^2-12-3{x}^2-x+4$

$=-x-8$

$解：原式=({a}^2+3a-a-3)+{a}^2-2a$

$=2{a}^2-3$

$将a=-1代入，得$

$原式=2×{(-1)}^2-3=-1$

$解：(a+1)(b+1)=ab+b+a+1=ab+(a+b)+1=4+5+1=10$

$解：({x}^{2}-mx+1)(x-2)={x}^{3}-m{x}^{2}+x-2{x}^{2}+2mx-2={x}^{3}-(m+2){x}^{2}+(2m+1)x-2$

$由题意得，m+2=0$

$解得，m=-2$

${x}^{2}-1$

${x}^{3}-1$

${x}^{4}-1$

${x}^{100}-1$

$解：(2)将x=2代入(x-1)({x}^{99}+{x}^{98}+···+x+1)={x}^{100}-1中，得$

$(2-1)({2}^{99}+{2}^{98}+···+2+1)={2}^{100}-1$

$所以{2}^{99}+{2}^{98}+···+2+1={2}^{100}-1$