Question

Bài 2: Tìm x nguyên

a) 20 – [4² + (x – 6)] = 90

b) (x + 3).(2x – 4) = 0

c) 1000:[30 + (2^x – 6)] = 3² + 4² và x ∈ N

Answer 1

\(2,\)

\(a,20-\left[4^2+\left(x-6\right)\right]=90\)

\(\Rightarrow20-16-x+6=90\)

\(\Rightarrow10-x=90\)

\(\Rightarrow x=-80\)

Vậy: \(x=-80\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2^x-6\right)=25\)

\(\Rightarrow24+2^x=40\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy: \(x=4\)

Answer 2

:))))