Tính bằng máy tính CASIO:
\(A=\dfrac{\left(1986^2-1992\right).\left(1986^2+3972-3\right).1987}{1983.1985.1987.1989}\)
\(B=26\left[\dfrac{3:\left(0,2-0,1\right)}{2,5.\left(0,8+1,2\right)}+\dfrac{\left(34,06-33,81\right).4}{6,84:\left(28,57-25,15\right)}\right]+\dfrac{2}{3}:\dfrac{4}{21}\)
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21 tháng 2 2020
Bài 11 : ta có :
\(\frac{\left(1986^2-1992\right)\left(1986^2+3972-3\right)1987}{1983.1985.1988.1989}\)
\(=\frac{\left(1986^2-3.1986+2.1986-6\right)\left(1986^2+2.1986+1-4\right)1987}{1983.1985.1988.1989}\)
\(=\frac{\left(1986-3\right)\left(1986+2\right)\left[\left(1986+1\right)^2-2^2\right]1987}{1983.1985.1988.1989}\)
\(=\frac{1983.1988\left(1987-2\right)\left(1987+2\right)1987}{1983.1988.1985.1989}\)
\(=\frac{1983.1985.1988.1989.1987}{1983.1985.1988.1989}=1987\)
28 tháng 9 2018
Đây là tính hợp lí ... mà câu a là 27,5 chứ không phải 2,75...
\(A=\dfrac{7,5-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{27,5-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{15}{2}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{55}{2}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{3}}\\ =\dfrac{3\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
8 tháng 9 2022
b: \(=26:\left[\dfrac{3:0.1}{2.5\cdot2}+\dfrac{0.25\cdot4}{2}\right]+\dfrac{2}{3}\cdot\dfrac{21}{4}\)
\(=26:\left[\dfrac{30}{5}+1\right]+\dfrac{42}{12}\)
\(=\dfrac{26}{7}+\dfrac{42}{12}=\dfrac{101}{14}\)
c: \(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{\left(\dfrac{1}{34}+\dfrac{33}{34}\right)}{\dfrac{1}{2}+\dfrac{9}{2}}=1:5=\dfrac{1}{5}\)
8 tháng 2 2021
Bài 1: Tìm x
a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)
Vậy: x=4
b) Ta có: \(\left|x\right|=-1.2\)
mà \(\left|x\right|\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
Bài 2: Tính
a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)
\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)
\(=-7.9\cdot10=-79\)
b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)
\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)
\(=3\cdot\dfrac{13}{3}=13\)
10 tháng 12 2022
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
NT
6 tháng 8 2021
a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)
c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)
