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電子課本網(wǎng) 第142頁(yè)

第142頁(yè)

信息發(fā)布者:
解:原式?$=-5\frac {3}{4}+7.4+4.75+2\frac {3}{5}$?
?$ =-5.75 + 7.4 + 4.75 + 2.6$?
?$ =(-5.75 + 4.75)+(7.4 + 2.6)$?
?$ =-1 + 10$?
?$ =9$?
解:原式?$=(-60)×(\frac {3}{4}+\frac {1}{10}-\frac {5}{6})$?
?$ =(-60)×\frac {3}{4}+(-60)×\frac {1}{10}-(-60)×\frac {5}{6}$?
?$ =-45 - 6 + 50$?
?$ =-51 + 50$?
?$ =-1$?
解:原式?$=-1^2 - 3×(-1)^3\div (-\frac {1}{2})^2$?
?$ =-1 - 3×(-1)\div \frac {1}{4}$?
?$ =-1 + 3×4$?
?$ =-1 + 12$?
?$ =11$?
解:?$(1)$?原式?$=(-2)3+33 $?
?$=-8+27 $?
?$=19 $?
解:?$(2)$?原式?$=(-2+3)×[(-2)2-3×(-2)+32] $?
?$=1×(4+6+9) $?
?$=19 $?
解:?$S_{陰影}=a2-b2$?
?$S_{陰影}=(a+b)(a-b)$?
則?$a2-b2=(a+b)(a-b)$?
解:?$(1)$?因?yàn)?$A = 2a^2+3ab - 2a - 1,$??$B=-a^2+ab + 1,$?
所以?$A + 2B=(2a^2+3ab - 2a - 1)+2(-a^2+ab + 1)$?
?$ =2a^2+3ab - 2a - 1 - 2a^2+2ab + 2$?
?$ =(2a^2-2a^2)+(3ab + 2ab)-2a+(-1 + 2)$?
?$ =5ab - 2a + 1$?
?$ $?當(dāng)?$a=-1,$??$b = 2$?時(shí),
原式?$=5×(-1)×2-2×(-1)+1=-10 + 2 + 1=-7$?
?$ (2)$?由?$ (1)$?知?$A + 2B=(5b - 2)a+1,$?
因?yàn)榇鷶?shù)式的值與?$a$?的取值無(wú)關(guān),
所以?$5b-2 = 0,$?
解得?$b=\frac {2}{5}$?
(1)解:因?yàn)?A = 2a^{2}+3ab - 2a - 1$,$B=-a^{2}+ab + 1$,所以$A + 2B=(2a^{2}+3ab - 2a - 1)+2(-a^{2}+ab + 1)$
$=2a^{2}+3ab - 2a - 1 - 2a^{2}+2ab + 2$
$=(2a^{2}-2a^{2})+(3ab + 2ab)-2a+(-1 + 2)$
$=5ab - 2a + 1$
當(dāng)$a=-1$,$b = 2$時(shí),原式$=5×(-1)×2-2×(-1)+1=-10 + 2 + 1=-7$
(2)解:由(1)知$A + 2B=(5b - 2)a+1$,因?yàn)榇鷶?shù)式的值與$a$的取值無(wú)關(guān),所以$5b-2 = 0$,解得$b=\frac{2}{5}$