toán 8 tập 1

D

deadguy

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T

thaolovely1412

1.
a.[TEX]a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)[(a^2-2ab+b^2)+ab]=(a+b)[(a-b)^2+ab][/TEX]
b.[TEX] (a^2+b^2)(c^2+d^2)[/TEX]
[TEX]=a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2+2abcd-2abcd[/TEX]
[TEX]=(a^2c^2+2abcd+b^2d^2)+(a^2d^2-2abcd+b^2c^2)=(ac+bd)^2+(ad-bc)^2[/TEX]
 
T

thienthannho118

Bài 2
N=2x-2x^2-5 = -2(x^2-x+5/2)=-2(x^2-x+1/4+9/4)=-2(x-1/2)^2-9/2 bé hơn hoặc bằng -9/2
 
T

thaolovely1412

2.tìm GTLN
[TEX]N=2x-2x^2-5=-2(x^2-x+2,5)=-2(x^2-x+0,25+2,25)=-2(x-0.5)^2-4,5 \leq -4,5[/TEX]
[TEX]\Rightarrow Max N=-4,5[/TEX]
 
B

buivanbao123

N=$2x-2x^{2}-5$
n=-$(2X^{2}-2X+5)$
N=$-2(x^{2}-x+5/2)$
N=$-2[(x-\dfrac{1}{2})^{2}+2)$
N=$-4-2(x-\dfrac{1}{2})^{2}$ \leq -4


@angle: sai rồi anh =))
 
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