第十章数的开方(单元测试·拔尖卷)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》8上多版本_2026《初中数学•举一反三系列》8上(华东师大)

                                                                           
                                                                       
第十章  数的开方·拔尖卷
 2024

Ⅰ 
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(10,10)37©4K(1+9)×9÷2+10=559(55÷3=18…1(\(10,10)cd<7·¸K 1(
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∴ab=30(
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∴ab−( 29)
2=30−29=1(
¬ab−( 29)
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O
a1= 1×2×3×4+1=1×(1+3)+1=5(
a2= 2×3×4×5+1=2×(2+3)+1=11(
a3= 3×4×5×6+1=3×(3+3)+1=19(
……(
µm°Ġs˜(an= n(n+1)(n+2)(n+3)+1=n(n+3)+1(
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∴
1
an+1=
1
(n+1)(n+2)= 1
n+1−
1
n+2(
∴
1
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1
a2+1+
1
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1
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∵a2=42Ăb3=−27
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ÿa=4(b=−3Û((a+b)2=(4−3)2=1(
ÿa=−4(b=−3Û((a+b)2=(−4−3)2=49(
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Ă|x−3|+ y+1+(z+2)2=0(
∴|x−3|=0( y+1=0((z+2)2=0(
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2
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*:;{´77(Ľb7tj−1<a<0(1<b<2(nqĻļrs"
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œrs"
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∵9<15<16(
∴3< 15<4(
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∴1< 15−2<2(
∴[ 15−2]=1Ë
∵3x+7
7
=4(
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7 <5(
t7≤x<
28
3 (
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(1)2×(−3)+ 25−|−4|+
3 27Ë
(2) 16+
3 −27
64×
−4
3
2
−|2−5|"
(1)−2
(2)5−5
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%1,先进行有理数乘法，算术平方根化简，化简绝对值，求立方根，然后通过有理数加减运算法则即可求
解．
（2）根据算术平方根与立方根化简，化简绝对值，然后通过有理数加减运算法则即可求解．
O%1,解：原式=−6+5−4+3=−2．
%2, 16+
3 −27
64×
−4
3
2
−|2−5|
=4−3
4 × 4
3 −( 5−2)
=4−1−5+2
=5−5．
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(1) 7−2
(2)2
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∴2< 7<3(
∴7<R7Ĵ*M 2('7Ĵ*M 7−2Ë
\M
 7−2Ë
%2,∵16<17<25(64<71<81(
∴4< 17<5(8< 71<9(
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∴17<'7Ĵ*Mx= 17−4( 71R7Ĵ*M y=8(
∴x+y−17= 17−4+8−17=4(
4<“—V@AK2"
∴x+y−17<“—V@AK2"
19"%8 *,%24-25 „…†‡·/ĿZŔ·¢£}¤,AB‡cŕ‡©Ŋ
x
15
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
15.9
x2
225
228.01
231.04
234.09
237.16
240.25
243.36
246.49
249.64
252.81
(1) 243.36=______( 23716=______( 0.023104=______Ë
(2) 25000Ŗ9R7ºŗŘřœ 2.3<ŘŚÌ%WXśĈą 0.01,Ë
(3)ª153< n<154(¬)¾Ŝŝ<R7n8______9"
(1)15.6Ë154Ë0.152
(2)158( 2.3≈1.52
(3)306
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wŞş«±nšœË
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O%1,
lcŠs˜(15.62=243.36(
∴ 
243.36=15.6Ë
∵ ∵ 1542=23716(
∴ 
23716=154Ë
∵ 0.1522=0.023104(
∴ 
0.023104=0.152"
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15.6Ë154Ë0.152Ë
%2,
∵ 
25000= 250×100=10 250(
Ŧ∵ 15.82=249.64,15.92=252.81(250−249.64=0.36,252.81−250=2.81,0.36<2.81(
∴ 
25000 158 ºŗŘË
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∵ 
231.04=15.2,2.3=231.04÷100(
∴ 
2.3= 231.04÷100=15.2÷10=1.52"
%3,
Ð153< n<154­šŢÛV@st1532<n<1542(
Óst1532=23409(1542=23716(
∴ n<ţÌŤťK23409<n<23716(
¬)¾Ŝŝ<R7n<97M23716−23409−1=3069"
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306"
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7Ð(8,25)<ū@ÐŪ7ÐM
(2,−5)[(−5,2)"
˜™–b
(1)ĔŗŬj7Ð(27,1.69)<ū@ÐŪ7Ð_______Ë
(2)ª7Ð(x,y)<!9ū@ÐŪ7ÐK −7, 3
2 (œx(y<ÌË
(3)ª7Ð(a,b)<!9ū@ÐŪ7ÐK(−4,−5)(œa+b<Ì"
(1)(3,−1.3)((−1.3,3)
(2)x=27
8 ,y=49
(3)−39Ĩ−109
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O%1,
3 27=m=3(−1.69=n=−1.3(
∴7Ð(27,1.69)<ū@ÐŪ7Ð(3,−1.3)((−1.3,3)Ë
%2,
∵
3 a=m(−b=n(¥7Ð(m,n)[(n,m)ŪM7Ð(a,b)<!Гū@ÐŪ7Д(
∴n≤0(
∵7Ð(x,y)<!9ū@ÐŪ7ÐK −7, 3
2 (
∴y=(−7)2=49(x= 3
2
3
=
27
8 Ë
%3,
ª
3 a=−4(−b=−5(
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¬a=−64(b=25(
∴a+b=−64+25=−39Ë 
ª
3 a=−5(−b=−4(
¬a=−125(b=16(
∴a+b=−125+16=−109Ë
a+b<ÌM−39Ĩ−109"
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