tính nhanh:
a) \(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
K
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C
23 tháng 2 2017
Ta tính các số mũ thành số hết
\(A=-1500-\left\{125.8-11.\left[49-40+8\left(121-121\right)\right]\right\}.2\)
\(A=-1500-\left\{1000-11.\left(9+0\right)\right\}.2\)
\(A=-1500-\left(1000-99\right).2\)
\(A=-1500-901.2\)
\(A=-1500-1802=-3302\)
EC
24 tháng 8 2016
\(A=10^3-\left\{-5^3.2^3-11.\left[x^2-5.2^3+\left(121-11^2\right)\right]\right\}\)
\(A=10^3-\left\{\left(-10\right)^3-11.\left[x^2-40\right]\right\}\)
\(A=10^3-\left\{\left(-10\right)^3-11x^2+440\right\}\)
\(A=10^3+10^3+11x^2-440\)
\(A=2000-440+11x^2\)
\(A=1560+11x^2\)
25 tháng 3 2017
\(1000-\left\{\left(-5\right)^3.\left(-2\right)^3-11.[7^2-5.2^3+8.\left(11^2-121\right)]\right\}\)
=\(1000-\left\{[\left(-5\right).\left(-2\right)]^3-11.[7^2-5.2^3+2^3.\left(11^2-11^2\right)]\right\}\)
= \(1000-\left\{1000-11.[7^2-2^3.\left(5+0\right)]\right\}\)
= \(1000-[1000-11.\left(7^2-2^3.5\right)\)
= \(1000-[1000-11.\left(49-40\right)]\)
= \(1000-\left(1000-11.9\right)\)
= \(1000-\left(1000-99\right)=1000-1000+99\)
= 0 + 99 = 99
21 tháng 5 2022
1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)
\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)
\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)
\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)
2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)
\(=1500-\left\{1000-11\cdot9\right\}\)
\(=1500-1000+99=599\)
T
23 tháng 8 2023
\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)
\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)
\(=5+\dfrac{19}{4}\)
\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)
\(------\)
\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)
\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)
\(=6-\dfrac{19}{7}\)
\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)
17 tháng 2 2022
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
2 tháng 10 2016
\(2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
Đặt : \(A=2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+0\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-40\right]\right\}\)
\(A=2017-\left\{25.4-11.9\right\}\)
\(A=2017-\left\{25.4-99\right\}\)
\(A=2017-\left\{100-99\right\}\)
\(A=2017-1=2016\)
Vậy A = 2016
\(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=1500-\left\{125.8-11.\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=1500-\left\{1000-11\left[49-40+8.0\right]\right\}\)
\(A=1500-\left\{1000-11.9\right\}\)
\(A=1500-\left\{1000-99\right\}\)
\(A=1500-901=599\)