Tìm số nguyên x :
\(720:\left[41-\left(2x+5\right)\right]=40\)
Tìm x biết:
\(25+3\left(x-8\right)=106\)
\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
Tìm x , biết
a, \(3^2\left(x+14\right)-5^2=5.4\)
b, \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
Ai nhanh minh tick
Help me !
Tìm x thuộc N , biết :
\(720:[41-\left(2x-5\right)]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=120\)
\(720:\left[41-\left(2x-5\right)\right]=^{ }2^3.5\)
\(720:\left[41-\left(2.x-5\right)\right]=2^3.5\)
\(720\div\left[41-\left(2x-5\right)\right]=2^3\times5\)
\((x-1)^3=125\)
\(2^{x+2}-2^x=96\)
\(\left(2x+1\right)^3=343\)
\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)\(16^x< 128^4\)
