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七年级数学解答题一般
题目
(1)(1)已知5a+23=3,3a+b=4,c\sqrt[3]{5a+2}=3,\sqrt{3a+b}=4,c11\sqrt{11}的整数部分,求a+b+ca+b+c的平方根.
(2)(2)已知:关于xx的多项式2(mx2x72)+4x2+3nx2(m{x}^{2}-x-\frac{7}{2})+4{x}^{2}+3nx的值与xx的取值无关.求3(2m23mn5m1)+6(m2+mn1)3(2m^{2}-3mn-5m-1)+6(-m^{2}+mn-1)的值.
知识点:相反数、绝对值(二)、平方根、算术平方根、代数式求值、倒数章节:第1章 有理数 / 1.2 有理数 / 1.2.4 绝对值

答案与解析

答案

(1)5a+23=33a+b=4\left(1\right)\because \sqrt[3]{5a+2}=3,\sqrt{3a+b}=4
5a+2=33=27\therefore 5a+2=3^{3}=273a+b=42=163a+b=4^{2}=16
联立方程组,得{5a+2=273a+b=16\left\{\begin{array}{l}5a+2=27\\ 3a+b=16\end{array}\right.
解得:{a=5b=1\left\{\begin{array}{l}a=5\\ b=1\end{array}\right.
3114\because 3<\sqrt{11}<4
c\because c11\sqrt{11}的整数部分,
c=3\therefore c=3
a+b+c=5+1+3=9\therefore a+b+c=5+1+3=9
a+b+c\therefore a+b+c的平方根为±9=±3±\sqrt{9}=±3
(2)2(mx2x72)+4x2+3nx(2)2(m{x}^{2}-x-\frac{7}{2})+4{x}^{2}+3nx
=2mx22x7+4x2+3nx=2mx^{2}-2x-7+4x^{2}+3nx
=(2m+4)x2+(3n2)x7=\left(2m+4\right)x^{2}+\left(3n-2\right)x-7
\because多项式2(mx2x72)+4x2+3nx2(m{x}^{2}-x-\frac{7}{2})+4{x}^{2}+3nx的值与xx的取值无关,
2m+4=0\therefore 2m+4=03n2=03n-2=0
解得:m=2n=23m=-2,n=\frac{2}{3}
3(2m23mn5m1)+6(m2+mn1)3(2m^{2}-3mn-5m-1)+6(-m^{2}+mn-1)
=6m29mn15m36m2+6mn6=6m^{2}-9mn-15m-3-6m^{2}+6mn-6
=3mn15m9=-3mn-15m-9
m=2n=23m=-2,n=\frac{2}{3}时,原式=3×(2)×2315×(2)9=25=-3×(-2)×\frac{2}{3}-15×(-2)-9=25.

解析

(1)5a+23=33a+b=4\left(1\right)\because \sqrt[3]{5a+2}=3,\sqrt{3a+b}=4
5a+2=33=27\therefore 5a+2=3^{3}=273a+b=42=163a+b=4^{2}=16
联立方程组,得{5a+2=273a+b=16\left\{\begin{array}{l}5a+2=27\\ 3a+b=16\end{array}\right.
解得:{a=5b=1\left\{\begin{array}{l}a=5\\ b=1\end{array}\right.
3114\because 3<\sqrt{11}<4
c\because c11\sqrt{11}的整数部分,
c=3\therefore c=3
a+b+c=5+1+3=9\therefore a+b+c=5+1+3=9
a+b+c\therefore a+b+c的平方根为±9=±3±\sqrt{9}=±3
(2)2(mx2x72)+4x2+3nx(2)2(m{x}^{2}-x-\frac{7}{2})+4{x}^{2}+3nx
=2mx22x7+4x2+3nx=2mx^{2}-2x-7+4x^{2}+3nx
=(2m+4)x2+(3n2)x7=\left(2m+4\right)x^{2}+\left(3n-2\right)x-7
\because多项式2(mx2x72)+4x2+3nx2(m{x}^{2}-x-\frac{7}{2})+4{x}^{2}+3nx的值与xx的取值无关,
2m+4=0\therefore 2m+4=03n2=03n-2=0
解得:m=2n=23m=-2,n=\frac{2}{3}
3(2m23mn5m1)+6(m2+mn1)3(2m^{2}-3mn-5m-1)+6(-m^{2}+mn-1)
=6m29mn15m36m2+6mn6=6m^{2}-9mn-15m-3-6m^{2}+6mn-6
=3mn15m9=-3mn-15m-9
m=2n=23m=-2,n=\frac{2}{3}时,原式=3×(2)×2315×(2)9=25=-3×(-2)×\frac{2}{3}-15×(-2)-9=25.

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