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对于任意实数$ x $，均能写成其整数部分$[x]$与小数部分$\{x\}$的和，即$ x = [x] + \{x\} $，其中$[x]$称为$ x $的整数部分，表示不超过$ x $的最大整数，$\{x\}$称为$ x $的小数部分. 如$ 7.12 = [7.12] + \{7.12\} = 7 + 0.12 $，$[7.12] = 7$，$\{7.12\} = 0.12$.
(1)当$ x ≥ 0, y ≥ 0 $时，以下命题中为真命题的是________（填序号）.
①$\{3\} = 0$；②$[\sqrt{15}] = 3$；③$\{x+1\} = \{x\} + 1$；④$[x+1] = [x] + 1$；⑤若$[x] = 4, [y] = 2$，则$[x+y]$所有可能的值为$ 6 $和$ 7 $；⑥$[x+y] ≤ [x] + [y]$.
(2)已知$ 5 ≤ \sqrt{a} ≤ 7, 4 ≤ \sqrt{b} ≤ 6 $，则$[\sqrt{a+b}]$可以是________.
(3)计算：$|\{6+\sqrt{5}\} - [6+\sqrt{5}]|$.
(4)计算：$\{\sqrt{16}-\sqrt{3}\} + [\sqrt{4}]$.
(5)计算：$[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + [\sqrt{4}] + [\sqrt{5}] + ··· + [\sqrt{47}] + [\sqrt{48}]$.
(6)当$ x ≥ 0 $时，解关于$ x $的方程$ 3x - 6 = 2[x+2] + \{x\} $.
(1)当$ x ≥ 0, y ≥ 0 $时，以下命题中为真命题的是________（填序号）.
①$\{3\} = 0$；②$[\sqrt{15}] = 3$；③$\{x+1\} = \{x\} + 1$；④$[x+1] = [x] + 1$；⑤若$[x] = 4, [y] = 2$，则$[x+y]$所有可能的值为$ 6 $和$ 7 $；⑥$[x+y] ≤ [x] + [y]$.
(2)已知$ 5 ≤ \sqrt{a} ≤ 7, 4 ≤ \sqrt{b} ≤ 6 $，则$[\sqrt{a+b}]$可以是________.
(3)计算：$|\{6+\sqrt{5}\} - [6+\sqrt{5}]|$.
(4)计算：$\{\sqrt{16}-\sqrt{3}\} + [\sqrt{4}]$.
(5)计算：$[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + [\sqrt{4}] + [\sqrt{5}] + ··· + [\sqrt{47}] + [\sqrt{48}]$.
(6)当$ x ≥ 0 $时，解关于$ x $的方程$ 3x - 6 = 2[x+2] + \{x\} $.
答案
(1)①②④⑤ 解析: $\{3\} = 3-3 = 0$,①正确;$\because \sqrt{9}＜\sqrt{15}＜\sqrt{16},\therefore 3＜\sqrt{15}＜4,\therefore [\sqrt{15}]=3$,②正确;$\because \{x+1\}$表示$x+1$的小数部分,$\therefore \{x+1\}=\{x\}$,③错误;$\because [x]$表示$x$的整数部分,$\therefore [x+1]=[x]+1$,④正确;$\because [x]\le x＜[x]+1,[y]\le y＜[y]+1,\therefore [x]+[y]\le x+y＜[x]+1+[y]+1,\therefore 6\le x+y＜8,\therefore 6\le [x+y]＜8,\therefore [x+y]$所有可能的值为6和7,⑤正确;当$x=1.5$,$y=2.6$时,$[x+y]=[4.1]=4$,$[x]+[y]=1+2=3$,$\therefore [x+y]＞[x]+[y]$,⑥错误.
(2)6,7,8,9 解析: $\because 5\le \sqrt{a}\le7,4\le\sqrt{b}\le6,\therefore 25\le a\le49,16\le b\le36,\therefore 41\le a+b\le85$,则$[\sqrt{a+b}]$可以是6,7,8,9.
(3)$\because 4＜5＜9,\therefore 2＜\sqrt{5}＜3,\therefore 8＜6+\sqrt{5}＜9,\therefore [6+\sqrt{5}]=8$,$\{6+\sqrt{5}\}=6+\sqrt{5}-8=\sqrt{5}-2$,$\therefore |\{6+\sqrt{5}\}-[6+\sqrt{5}]|=|\sqrt{5}-2-8|=|\sqrt{5}-10|=10-\sqrt{5}$.
(4)$\because 1＜\sqrt{3}＜2,\sqrt{16}=4,\therefore 4-2＜\sqrt{16}-\sqrt{3}＜4-1$,即$2＜\sqrt{16}-\sqrt{3}＜3$,$\{\sqrt{16}-\sqrt{3}\}=4-\sqrt{3}-2=2-\sqrt{3}$,$[\sqrt{4}]=2$,则$\{\sqrt{16}-\sqrt{3}\}+[\sqrt{4}]=2-\sqrt{3}+2=4-\sqrt{3}$.
(5)$\because \sqrt{1}=1,\sqrt{4}=2,\sqrt{9}=3,\sqrt{16}=4,\sqrt{25}=5,\sqrt{36}=6,\sqrt{49}=7,\therefore [\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+[\sqrt{4}]+\dots+[\sqrt{47}]+[\sqrt{48}]=1×(4-1)+2×(9-4)+3×(16-9)+4×(25-16)+5×(36-25)+6×(49-36)=3+10+21+36+55+78=203$.
(6)$\because x\ge0,\therefore 2[x+2]+\{x\}=2([x]+2)+\{x\}=2[x]+4+\{x\}=2x-\{x\}+4$,$\therefore 3x-6=2x-\{x\}+4$,$\therefore x+\{x\}=10$.$\because \{x\}$是$x$的小数部分,当$\{x\}=0$时,$x=10$;当$\{x\}\ne0$时,$[x]=9$,$\therefore x+\{x\}=[x]+2\{x\}=9+2\{x\}=10$,可得$\{x\}=0.5$,$\therefore x=9+0.5=9.5$.综上可得$x=10$或$x=9.5$.
(2)6,7,8,9 解析: $\because 5\le \sqrt{a}\le7,4\le\sqrt{b}\le6,\therefore 25\le a\le49,16\le b\le36,\therefore 41\le a+b\le85$,则$[\sqrt{a+b}]$可以是6,7,8,9.
(3)$\because 4＜5＜9,\therefore 2＜\sqrt{5}＜3,\therefore 8＜6+\sqrt{5}＜9,\therefore [6+\sqrt{5}]=8$,$\{6+\sqrt{5}\}=6+\sqrt{5}-8=\sqrt{5}-2$,$\therefore |\{6+\sqrt{5}\}-[6+\sqrt{5}]|=|\sqrt{5}-2-8|=|\sqrt{5}-10|=10-\sqrt{5}$.
(4)$\because 1＜\sqrt{3}＜2,\sqrt{16}=4,\therefore 4-2＜\sqrt{16}-\sqrt{3}＜4-1$,即$2＜\sqrt{16}-\sqrt{3}＜3$,$\{\sqrt{16}-\sqrt{3}\}=4-\sqrt{3}-2=2-\sqrt{3}$,$[\sqrt{4}]=2$,则$\{\sqrt{16}-\sqrt{3}\}+[\sqrt{4}]=2-\sqrt{3}+2=4-\sqrt{3}$.
(5)$\because \sqrt{1}=1,\sqrt{4}=2,\sqrt{9}=3,\sqrt{16}=4,\sqrt{25}=5,\sqrt{36}=6,\sqrt{49}=7,\therefore [\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+[\sqrt{4}]+\dots+[\sqrt{47}]+[\sqrt{48}]=1×(4-1)+2×(9-4)+3×(16-9)+4×(25-16)+5×(36-25)+6×(49-36)=3+10+21+36+55+78=203$.
(6)$\because x\ge0,\therefore 2[x+2]+\{x\}=2([x]+2)+\{x\}=2[x]+4+\{x\}=2x-\{x\}+4$,$\therefore 3x-6=2x-\{x\}+4$,$\therefore x+\{x\}=10$.$\because \{x\}$是$x$的小数部分,当$\{x\}=0$时,$x=10$;当$\{x\}\ne0$时,$[x]=9$,$\therefore x+\{x\}=[x]+2\{x\}=9+2\{x\}=10$,可得$\{x\}=0.5$,$\therefore x=9+0.5=9.5$.综上可得$x=10$或$x=9.5$.
