解:[10n+(n+2)]²-138=10(n+2)+n
(10n+n+2)²-138=10n+20+n
(11n+2)²-138=11n+20
121n²+44n+4-138=11n+20
121n²+44n-11n+4-138-20=0
121n²+33n-154=0,两边同时除以11得:
11n²+3n-14=0,十字相乘法分解,得:
(n-1)(11n+14)=0
n-1=0,11n+14=0
n1=1,n2=-14/11
[10n+(n+2)]^2-138=10(n+2)+n
(11n+2)^2-11n-158=0
121n^2+33n-154=0
11n^2+3n-14=0
(11n+14)(n-1)=0
11n+14=0
n=-14/11
n-1=0
n=1
【10n+(n+2)】^2-138=10(n+2)+n,公式化简得:121n^2+33n-154=0
=>(11n-11)(11n+14)=0 ==>n=1或者-14/11