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Bài 2:Đặt n+5=\(m^2\);n+30=\(k^2\) (m,k\(\in N\)* ; m<k)

Ta có:\(k^2-m^2=n+30-n-5=25\)

Vì k>m>0 \(\Rightarrow\) k-m \(\ne\) k+m và k+m > k-m > 0

\(\Rightarrow\left(k-m\right)\left(k+m\right)=1.25\)

\(\Rightarrow\left\{{}\begin{matrix}k-m=1\\k+m=25\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}k=13\\m=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n+5=12^2=144\\n+30=13^2=169\end{matrix}\right.\)\(\Rightarrow n=139\)

-6.(1 - x) + 5.(2 - 5.x) = 42

=> -6 + 6x + 10 - 25x = 42

=> 4 -19x = 42

=> 19x = 4 - 42

=> 19x = -38

=> x = -38 : 19

=> x = -2