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LuyÖn thi trªn m¹ng
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DÊu b»ng x¶y ra khi a = b = c. VËy ABC ®Òu.
a
b) Û b + c =
+
3bsinC Û
2
1
1
sinB + sinC = sinA + 3sinBsinC Û sinB + sinC = sin(B + C) + 3sinBsinC
2
2
1
Û sinB + sinC = [sinBcosC + sinCcosB] + 3sinBsinC
2
cosC
2
3
cosB
2
3 π
sinB = 0 Û sinB 1 - sin(C + ) + sinC 1 - sin(B + ) = 0
6
2
6
π
ÛsinB 1-
-
sinC + sinC 1-
-
2
π
π
sin(C + ) =1
C =
3
3
⇔
⇔
π
sin(B+ ) =1
6
π
B =
3
2
) §Æt tgx + cotgx = t(|t| ³ 2) th× sÏ cã:
2
tg x + cotg x = (tgx + cotgx) - 2 = t - 2;
2
2
2
3
3
3
3
tg x + cotg x =(tgx + cotgx) - 3tgxcotgx (tgx + cotgx) = t - 3t.
2
3
VËy ta cã phû¬ng tr×nh: t + (t - 2) + (t - 3t) = 6
3
hay t + t - 2t - 8 = 0 Û (t - 2) (t + 3t + 4) = 0 Û t = 2.
2
2
π
Sau ®ã gi¶i phû¬ng tr×nh: tgx + cotgx = 2 sÏ ®ûîc mét hä nghiÖm lµ: x =
C©u III. 1) ViÕt l¹i phû¬ng tr×nh ®· cho:
+ kπ (k Î Z).
4
2
x - 2x + 5 = - 4cos(ax + b) Û (x - 1) + 4 = - 4cos(ax + b) .(1)
2
2
Ta cã:(x - 1) + 4 ³ 4 ³ - 4cos(ax + b).
V× thÕ x lµ nghiÖm cña (1) khi vµ chØ khi x lµ nghiÖm cña hÖ: