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8. (2023·山东改编)若$\triangle ABC的三边长a,b,c满足(a-b)^{2}+\sqrt {2a-b-3}+|c-3\sqrt {2}|= 0$,试确定$\triangle ABC$的形状,并说明理由.
答案
$ \because (a - b)^2 + \sqrt{2a - b - 3} + |c - 3\sqrt{2}| = 0 $，
又 $ \left\{ \begin{array} { l } { ( a - b ) ^ { 2 } \geq 0, } \\ { \sqrt { 2 a - b - 3 } \geq 0, } \\ { | c - 3 \sqrt { 2 } | \geq 0, } \end{array} \right. $ $ \therefore \left\{ \begin{array} { l } { ( a - b ) ^ { 2 } = 0, } \\ { \sqrt { 2 a - b - 3 } = 0, } \\ { | c - 3 \sqrt { 2 } | = 0, } \end{array} \right. $ $ \therefore \left\{ \begin{array} { l } { a - b = 0, } \\ { 2 a - b - 3 = 0, } \\ { c - 3 \sqrt { 2 } = 0, } \end{array} \right. $ 解得 $ \left\{ \begin{array} { l } { a = 3, } \\ { b = 3, } \\ { c = 3 \sqrt { 2 }, } \end{array} \right. $
$ \therefore a ^ { 2 } + b ^ { 2 } = c ^ { 2 } $，且 $ a = b $，
$ \therefore \triangle ABC $ 为等腰直角三角形.
又 $ \left\{ \begin{array} { l } { ( a - b ) ^ { 2 } \geq 0, } \\ { \sqrt { 2 a - b - 3 } \geq 0, } \\ { | c - 3 \sqrt { 2 } | \geq 0, } \end{array} \right. $ $ \therefore \left\{ \begin{array} { l } { ( a - b ) ^ { 2 } = 0, } \\ { \sqrt { 2 a - b - 3 } = 0, } \\ { | c - 3 \sqrt { 2 } | = 0, } \end{array} \right. $ $ \therefore \left\{ \begin{array} { l } { a - b = 0, } \\ { 2 a - b - 3 = 0, } \\ { c - 3 \sqrt { 2 } = 0, } \end{array} \right. $ 解得 $ \left\{ \begin{array} { l } { a = 3, } \\ { b = 3, } \\ { c = 3 \sqrt { 2 }, } \end{array} \right. $
$ \therefore a ^ { 2 } + b ^ { 2 } = c ^ { 2 } $，且 $ a = b $，
$ \therefore \triangle ABC $ 为等腰直角三角形.
9. 甲、乙两人计算$a+\sqrt {1-2a+a^{2}}$的值,当$a= 5$时,得出不同的答案. 甲的解答是$a+\sqrt {1-2a+a^{2}}= a+\sqrt {(1-a)^{2}}= a+1-a= 1$,乙的解答是$a+\sqrt {1-2a+a^{2}}= a+\sqrt {(a-1)^{2}}= a+a-1= 2a-1= 2×5-1= 9$. 谁的解答是正确的? 错误的解答错在哪里? 为什么?
答案
乙的解答是正确的. 当 $ a = 5 $ 时，$ \sqrt { 1 - 2 a + a ^ { 2 } } = \sqrt { ( 1 - a ) ^ { 2 } } = a - 1 $.
