\(\left(\dfrac{1}{9}\right)^{25}\cdot3^{50}=\dfrac{1^{25}}{\left(3^2\right)^{25}}\cdot3^{50}=\dfrac{1^{25}}{3^{50}}\cdot3^{50}=1\)
\(\left(\dfrac{1}{9}\right)^{25}\cdot3^{50}=\dfrac{1^{25}}{\left(3^2\right)^{25}}\cdot3^{50}=\dfrac{1^{25}}{3^{50}}\cdot3^{50}=1\)
\((\dfrac{1}{3})^{50}\times(-9)^{25}-\dfrac{2}{3}\div4\)
8)\(\left(\dfrac{1}{3}\right)^{50}.\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
o) \(\dfrac{\left(-1\right)^6.3^5.4^3}{9^2.2^5}\)
s) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
t) \(\sqrt{\dfrac{4}{9}}\) - \(\dfrac{1}{2}\): \(\left|\dfrac{-2}{3}\right|\)
Mg giải gấp giúp mình với
cho \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\) và \(2x^3-1=15\) tính \(B=x+y+z\)
Câu 1: Thực hiện phép tính
a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)
b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)
Câu 2: Tìm x
\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)
a,2.(\(\dfrac{1}{4}\)+x)\(^3\)=(\(-\dfrac{27}{4}\))
b,(x+\(\dfrac{1}{2}\))\(^3\):3=\(\dfrac{-1}{81}\)
c,(\(\dfrac{2}{3}\)-x)\(^2\)=1:\(\dfrac{4}{9}\)
d,(2x-\(\dfrac{1}{5}\))\(^2\)+\(\dfrac{16}{25}\)=1
e,(\(\dfrac{2}{5}\)-3x)\(^2\)-\(\dfrac{1}{5}\)=\(\dfrac{4}{25}\)
Bài 1: Tính giá trị của biểu thức sau
A=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)
B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0
Mong mn giúp đỡ
a \(\dfrac{-4}{7}\) - \(\dfrac{5}{13}\) x \(\dfrac{-39}{25}\) + \(\dfrac{-1}{42}\) : \(\dfrac{-5}{6}\)
b \(\dfrac{2}{9}\) x [\(\dfrac{4}{45}\): ( \(\dfrac{1}{5}\) - \(\dfrac{2}{15}\)) + 1\(\dfrac{2}{3}\)] - \(\dfrac{-5}{27}\)
bài 6: tính :
\(\dfrac{10^9.\left(-81\right)^{10}}{\left(-8\right)^4.25^5.9^{10}}\)
b,\(\dfrac{9^4.\left(-4\right)^5.25^3}{8^3,\left(-27\right)^2.5^7}\)
c,\(\dfrac{3^{186}.\left(-25\right)^{50}}{\left(-15\right)^{100}.27^{29}}\)