解:(1)原式=-1×3
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+3
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+2×
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,
=-3
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+3
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+1,
=1,
(2)去分母得:2x(x-2)-(x-3)(x+1)=(x-3)(x-2),
去括號得:2x
2-2x-x
2+2x+3=x
2-5x+6.
移項得:2x
2-2x+2x-x
2-x
2+5x=6-3,
合并同類項得:5x=3,
把x的系數(shù)化為1:x=
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,
檢驗:把x=
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代入最簡公分母(x-3)(x-2)≠0,
∴原分式方程的解為:x=
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,
(3)由①得:x≤3,
由②得:x>1,
∴不等式組的解集為:3≥x>1,
∴整數(shù)解為:2,3.
分析:(1)此題涉及到了二次根式的化簡,乘方,絕對值,特殊角的三角函數(shù)值,首先根據(jù)各知識點(diǎn)計算,然后再算乘法,最后計算加減法即可.
(2)首先乘以最簡公分母去分母,然后去括號,移項,合并同類項,把x的系數(shù)化為1,最后一定要檢驗.
(3)分別解出兩個不等式的解集,然后再取公共解集再寫出整數(shù)解即可.
點(diǎn)評:此題主要考查了不等式組的解法,分式方程的解法,以及二次根式的化簡,乘方,絕對值,特殊角的三角函數(shù)值,實數(shù)運(yùn)算,解題的關(guān)鍵是準(zhǔn)確把握各個知識點(diǎn).