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10 tháng 4 2019

\(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)

\(A=1500-\left\{125.8-11.\left[49-5.8+8\left(121-121\right)\right]\right\}\)

\(A=1500-\left\{1000-11\left[49-40+8.0\right]\right\}\)

\(A=1500-\left\{1000-11.9\right\}\)

\(A=1500-\left\{1000-99\right\}\)

\(A=1500-901=599\)

23 tháng 2 2017

Ta tính các số mũ thành số hết

\(A=-1500-\left\{125.8-11.\left[49-40+8\left(121-121\right)\right]\right\}.2\)

\(A=-1500-\left\{1000-11.\left(9+0\right)\right\}.2\)

\(A=-1500-\left(1000-99\right).2\)

\(A=-1500-901.2\)

\(A=-1500-1802=-3302\)

24 tháng 2 2017

sai rồi nhé Hậu Trần Công đề bài là -2 chứ ko phải là 2

ok

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

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: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)

\(=1500-\left\{1000-11\cdot9\right\}\)

\(=1500-1000+99=599\)

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

 

27 tháng 2 2022

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)

27 tháng 2 2022

Viết lại phần d) đc 0 ạ=((