Lời giải:Ta có:
\(B= \sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(= \sqrt{16(x+1)}-\sqrt{9(x+1)}+\sqrt{4(x+1)}+\sqrt{x+1}\)
\(= \sqrt{4^2(x+1)}-\sqrt{3^2(x+1)}+\sqrt{2^2(x+1)}+\sqrt{x+1}\)
\(= 4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=(4-3+2+1)\sqrt{x+1}\)
\(=4\sqrt{x+1}.\)
\(B = 16 \Leftrightarrow 4\sqrt {x + 1} = 16\)
\(\eqalign{
& \Leftrightarrow \sqrt {x + 1} = {{16} \over 4} \cr
& \Leftrightarrow \sqrt {x + 1} = 4 \cr
& \Leftrightarrow {\left( {\sqrt {x + 1} } \right)^2} = {4^2} \cr
& \Leftrightarrow x + 1 = 16 \cr
& \Leftrightarrow x = 16 - 1 \cr
& \Leftrightarrow x = 15(\text{thỏa mãn}\,x\ge -1) \cr} \)
Vậy với \(x=15\) thì \(B=16\).
