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a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
a. Chu vi là \(\left(12+5\right).2=34\left(m\right)\)
Diện tích là \(12.5=60\left(m^2\right)=600000\left(cm^2\right)\)
b. Cần lát \(600000:\left(40.40\right)=375\) viên gạch
\(a,-\left(m+n-k\right)+\left(m-k\right)-\left(-m+n\right)\\ =-m-n+k+m-k+m-n\\ =\left(-m+m+m\right)+\left(-n-n\right)+\left(k-k\right)\\ =m-2n\)
\(b,\left(x-y\right)-\left(x+y\right)-\left(2x-3y\right)\\ =x-y-x-y-2x+3y\\ =\left(x-x-2x\right)+\left(-y-y+3y\right)\\ =-2x+y\)
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
=> \(\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}\)
Bài làm
12 . 59 + 59 . 137 - 59 . 49
= 59 . ( 12 + 137 - 49 )
= 59 . 100
= 5900
~ Dấu " . " là dấu nhân nha. ~
# Học tốt #
3d:
20<x<45
x chia 4 dư 1 nên x-1 thuộc B(4)
=>\(x-1\in\left\{0;4;...;44;48\right\}\)
=>\(x\in\left\{1;5;...;45;49\right\}\)
mà 20<x<45
nên x thuộc {21;26;31;35;41}
4:
a: A={x∈N|51<=x<=127}
b: B={x∈N|100<=x<=999}
c: C={x∈N|x=7k+5; 0<=k<=8}
1)Ta có:
\(\dfrac{999}{1000}+\dfrac{998}{1000}+\dfrac{997}{1000}+...+\dfrac{1}{1000}\)
=\(\dfrac{999+998+997+..+1}{1000}\)
= \(\dfrac{\left(999+1\right)+\left(998+2\right)+...+\left(499+501\right)+500}{1000}\)
=\(\dfrac{1000+1000+...+1000+500}{1000}\)
= \(\dfrac{1000.499+500}{1000}\)
= \(\dfrac{1000.499}{1000}+\dfrac{500}{1000}\)
= \(499+\dfrac{1}{2}\)
=499,5