分析:(1)本題應(yīng)對(duì)式中每個(gè)數(shù)化簡(jiǎn),然后去括號(hào)、合并同類項(xiàng)即可;
(2)本題應(yīng)對(duì)原式進(jìn)行配方,然后再開(kāi)方、移項(xiàng)即可得出x的值;
(3)本題可先對(duì)方程移項(xiàng),消去(x-5),再去括號(hào)、合并同類項(xiàng)即可;
(4)本題只要把圖書(shū)的三角函數(shù)值代入化簡(jiǎn)即可.
解答:解:(1)(
+
)-(
-
)
=(4
+2
)-(2
-
)
=4
+2
-2
+
=2
+3
;
(2)移項(xiàng),得x
2-2x=4
配方x
2-2x+1=4+1
(x-1)
2=5
由此可得x-1=±
∴x
1=1+
,x
2=1-
;
(3)原式=3(x-5)
2+2(x-5)=0
即3(x-5)+2=0
3x-15+2=0
x=
∴x
1=5,x
2=
;
(4)原式=3×(
)
2+2|
-1|
=3×
+2(1-
)
=1+2-
=3-
.
點(diǎn)評(píng):本題考查了二次根式的化簡(jiǎn)、一元二次方程的計(jì)算和特殊三角函數(shù)值的運(yùn)用.解此類題目時(shí)要對(duì)考點(diǎn)一一進(jìn)行分析再作答.