Toán Lớp 8: a ²-2a+b ²+4b+4c ²-4c+6=0
tìm a,b,c
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Toán Lớp 8: a ²-2a+b ²+4b+4c ²-4c+6=0
tìm a,b,c
Comments ( 2 )
$a^2$ − 2a + $b^2$ + 4b + 4$c^2$ − 4c + 6 = 0
⇒ ($a^2$ – 2a + 1)+ ($b^2$ + 4b + 4)+(4$c^2$ − 4c + 1) = 0
⇒ $(a – 1)^2$ + $(b + 2)^2$ + $(2c – 1)^2$ = 0
Do
$(a – 1)^2$ ≥ 0
$(b + 2)^2$ ≥ 0
$(2c – 1)^2$ ≥ 0
⇒ $(a – 1)^2$ ≥ 0 + $(b + 2)^2$ + $(2c – 1)^2$ ≥ 0
Dấu “=” xảy ra khi hoặc chỉ khi :
a−1 = 0
b+2=0
2c−1=0
$\Rightarrow$ a=1
b=−2
c= $\dfrac{1}{2}$
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a^2-2a+b^2+4b+4c^2-4c+6=0
-> (a^2 -2a+1) + (b^2+4b +4) + (4c^2 – 4c+1)=0
->(a-1)^2+ (b+2)^2 + (2c-1)^2=0
Do (a-1)^2\ge 0,(b+2)^2\ge 0,(2c-1)^2\ge 0
->(a-1)^2 + (b+2)^2 + (2c-1)^2\ge 0
Dấu “=” xảy ra khi : a-1=0, b+2=0,2c-1=0
<=>a=1,b=-2, c=1/2
Vậy a=1,b=-2,c=1/2