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a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
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\(x^3+\dfrac{3}{4}x+\dfrac{3}{2}x^2+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
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a: \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)
\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)
d: \(=\left(2x+5y\right)^3\)
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\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
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c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
a, Nhớ t/c này nhé ! \(\left(a-b\right)^2=\left(b-a\right)^2\)
\(\left(x-3\right)=\left(3-x\right)^2\Leftrightarrow\left(x-3\right)=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(1-x+3\right)=0\Leftrightarrow x=3;x=4\)
b, viết rõ đề ib mình giải tiếp nhé
cái 2 4 8 64 kia là gì vậy ?:))
a) ( x - 3 ) = ( 3 - x )2
<=> ( x - 3 ) - ( x - 3 )2 = 0
<=> ( x - 3 )[ 1 - ( x - 3 ) ] = 0
<=> ( x - 3 )( 1 - x + 3 ) = 0
<=> ( x - 3 )( 4 - x ) = 0
<=> x - 3 = 0 hoặc 4 - x = 0
<=> x = 3 hoặc x = 4
Vậy S = { 3 ; 4 }
b) x3 + 3x2 + 3x + 1 = 1
<=> ( x + 1 )3 - 1 = 0
<=> ( x + 1 - 1 )[ ( x + 1 )2 + x + 1 + 1 ) = 0
<=> x( x2 + 2x + 1 + x + 2 ) = 0
<=> x( x2 + 3x + 3 ) = 0
<=> x = 0 [ x2 + 3x + 3 = ( x2 + 3x + 9/4 ) + 3/4 = ( x + 3/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x ]
Vậy S = { 0 }