Toán Lớp 6: 1)Tính giá trị của biểu thức:
a) 3².10³ -[13² – (5².4+2².15)] .10³
b) 2⁴.5-[131-(13-4)²]
c)2/3+5/6 : 5 – 1/18.(-3)²
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Toán Lớp 6: 1)Tính giá trị của biểu thức:
a) 3².10³ -[13² – (5².4+2².15)] .10³
b) 2⁴.5-[131-(13-4)²]
c)2/3+5/6 : 5 – 1/18.(-3)²
Comments ( 1 )
a){3^2}{.10^3} – \left[ {{{13}^2} – \left( {{5^2}.4 + {2^2}.15} \right)} \right]{.10^3}\\
= {9.10^3} – \left( {169 – 4.\left( {25 + 15} \right)} \right){.10^3}\\
= {9.10^3} – \left( {169 – 4.35} \right){.10^3}\\
= {9.10^3} – {29.10^3}\\
= {10^3}.\left( {9 – 29} \right)\\
= 1000.\left( { – 20} \right)\\
= – 20000\\
b){2^4}.5 – \left( {131 – {{\left( {13 – 4} \right)}^2}} \right)\\
= 16.5 – \left( {131 – {9^2}} \right)\\
= 80 – \left( {131 – 81} \right)\\
= 80 – 50\\
= 30\\
c)\dfrac{2}{3} + \dfrac{5}{6}:5 – \dfrac{1}{{18}}.{\left( { – 3} \right)^2}\\
= \dfrac{2}{3} + \dfrac{1}{6} – \dfrac{1}{{18}}.9\\
= \dfrac{2}{3} + \dfrac{1}{6} – \dfrac{1}{2}\\
= \dfrac{{2.2 + 1 – 3}}{6}\\
= \dfrac{2}{6}\\
= \dfrac{1}{3}
\end{array}$