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苏科版七年级下册数学补充习题第58页答案

2024-12-21 22:28| 查看: 2

 (2)
(4.5×10³)(2×10³)(3×10);

=(4.5×2×3)×(10×10³×10)

=27×10¹² 

=2.7×10¹³


(3)

=-16x²y³+20x²y³ 

=4a²y³


(4)

=2x²-xy+6xy-3y² 

=2x²+5xy-3y²


(5)

=3x³-6x²-3x-2a³+6x² 

=x³-3x


13.

解:原式=3(x²+x-20)-2(x²-36)

=3x²+3x-60-2x²+72

=x²+3x+12

将x=-
1
2
代入得
x²+3x+12=(-
1
2
)²+3×(-
1
2
)+12=
43
4



14.把下列各式分解因式
(1)

=[4+(2a+3b)|[4-(2a+3b)]

=(4+2a+3b)(4-2a-36) 


(2)

     

=-(36x²-12xy+y²) 

=-(6x-y)²

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