\(2^7+\left(x+19\right)=\dfrac{900}{6}\)
\(\Rightarrow128+x+19=150\)
\(\Rightarrow x=150-128-19\)
\(\Rightarrow x=3\)
\(2^7+\left(x+19\right)=900:6\)
\(128+\left(x+19\right)=150\)
\(\left(x+19\right)=150-128\)
\(x+19=22\)
\(x=22-19\)
\(x=3\)
\(Vậy\)\(x=3\)
`2^{7}+(x+19)=900:6`
`->128+x+19=150`
`->x+147=150`
`->x=150-147`
`->x=3` (thỏa mãn yêu cầu đề bài)
Ta có: \(2^7+\left(x+19\right)=900:6\)
\(\Leftrightarrow x+19+128=150\)
hay x=3
