Toán tìm x

T

tranvanhung7997

a, $(x - 2)^3 - x^3 + 6x^2 = 7$
<=> $x^3 - 6x^2 + 12x - 8 - x^3 + 6x^2 = 7$
<=> $12x = 15$
<=> $x = \dfrac{5}{4}$
 
L

linhgiateo

tìm x

b, (x-3)^2 - (x-3)(x^2+3x+9) + 6(x+1)^2 - 7x^2 + x^3
=> x^2 - 6x + 9 -x^3-3x^2-9x + 3x^2+9x+27 + 6x^2+12x+6-7x^2+x^3=-33
=> 6x+42=-33 => x= (-33-42):6=-12,5
 
V

vipboycodon

c)[TEX](x-1)^3-(x+3)(x^2-3x+9)+3(x-2)(x+2)=2[/TEX]
\Leftrightarrow [TEX]x^3-3x^2+3x-1-(x^3-27)+3(x^2-4)=2[/TEX]
\Leftrightarrow [TEX]x^3-3x^2+3x-1-x^3+27+3x^2-12=2[/TEX]
\Leftrightarrow 3x+14=2
\Leftrightarrow 3x=-12
\Leftrightarrow x=-4

d)[TEX]x(x-5)(x+5)-(x^2-2x+4)(x+2)=3[/TEX]
\Leftrightarrow [TEX]x(x^2-25)-(x^3-8)=3[/TEX]
\Leftrightarrow [TEX]x^3-25x-x^3+8=3[/TEX]
\Leftrightarrow -25x=3-8
\Leftrightarrow -25x=-5
\Leftrightarrow [TEX]x=\frac{1}{5}[/TEX]
 
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S

supperdragon9510

c)[TEX](x-1)^3-(x+3)(x^2-3x+9)+3(x-2)(x+2)=2[/TEX]
\Leftrightarrow [TEX]x^3-3x^2+3x-1-(x^3-27)+3(x^2-4)=2[/TEX]
\Leftrightarrow [TEX]x^3-3x^2+3x-1-x^3+27+3x^2-12=2[/TEX]
\Leftrightarrow 3x+14=2
\Leftrightarrow 3x=-12
\Leftrightarrow x=-4

d)[TEX]x(x-5)(x+5)-(x^2-2x+4)(x+2)=3[/TEX]
\Leftrightarrow [TEX]x(x^2-25)-(x^3-9)=3[/TEX]
\Leftrightarrow [TEX]x^3-25x-x^3+9=3[/TEX]
\Leftrightarrow -25x=3-9
\Leftrightarrow -25x=-6
\Leftrightarrow [TEX]x=\frac{6}{25}[/TEX]
hình như sai rồi bạn ơi,sao $2^3$ lại thành 9 đc?xem lại đi nhé:)
 
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