\(6\div2\left(1+2\right)\)
\(=6\div2.3\)
\(=3.3\)
\(=9\)
\(6\div2\left(1+2\right)=6\div2.3\\ =6\div6\\ =1 \)
\(6\div2\times\left(1+2\right)=3\times\left(1+2\right)\\ =3\times3\\ =9\)
\(\dfrac{9}{48}\times\left(-2,4\right)+\left(\dfrac{1}{4}+\dfrac{13}{20}\right)\div2\)
\(\left(0,66-0,012\div0,2\right)\div\left(1-1\frac{4}{7}\times0,4\right)=2\frac{9}{13}\times x\div\left(3,125-5,6\div2\frac{2}{3}\right)\)
Cho mình xin cả cách làm nữa
A= \(-1,6\div\left(1+\frac{2}{3}\right)\)
B=\(1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right)\div2\frac{1}{5}\)
\(\left(2^{31}\div2\right)-\left(1+3+3^2+3^3+.....+3^{20}\right)\)
Tìm số tự nhiên n , biết: \(2\times\left(3^{2010}-3\right)\div2+3=3^{5n+5}\)
\(2020^{2020}\times\left(7^{10}\times7^8-3\times2^4-2^{2020}\div2^{2020}\right)\)
\(1\frac{3}{7}-\frac{4}{5}\) b)\(\frac{6}{13}\times\frac{-3}{10}+\frac{2}{5}\times\frac{4}{13}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2018}{2019}\)
Các bạn ơi giúp mk với
cho:
\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)
và \(B=\left(\frac{3}{2}\right)^{2013}\div2\)
tính B - A
\(2020^{2020}\times\left(7^{10}\times7^8-3\times2^4-2^{2020}\div2^{2020}\right)\)
Tìm x:
\(\frac{1}{2}\times x-\frac{3}{5}=\frac{-4}{5}\) B)\(\left(x-\frac{2}{3}\right)\div\frac{-3}{7}=\frac{-9}{14}\)
