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6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
Bài 2 :
a. A = 2 ( x3 + y3 ) - 3 ( x2 + y2 ) với x + y = 1
=> A = 2 ( x + y ) ( x2 - xy + y2 ) - 3 [ ( x + y )2 - 2xy ]
=> A = 2 [ ( x + y )2 - 3xy ] - 3 ( 1 - 2xy )
=> A = 2 ( 1 - 3xy ) - 3 + 6xy
=> A = 2 - 6xy - 3 + 6xy
=> A = - 1
B = x3 + y3 + 3xy với x + y = 1
=> B = ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
=> B = ( x + y )3 - 3xy ( x + y - 1 )
=> B = 13 - 3xy . 0
=> B = 1
Bài 1.
a) ( x - 1 )3 + ( 2 - x )( 4 + 2x + x2 ) + 3x( x + 2 ) = 16
<=> x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x = 16
<=> 9x + 7 = 16
<=> 9x = 9
<=> x = 1
b) ( x + 2 )( x2 - 2x + 4 ) - x( x2 - 2 ) = 15
<=> x3 + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 7
<=> x = 7/2
c) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 15
<=> ( x - 3 )[ ( x - 3 )2 - ( x2 + 3x + 9 ) + 9( x2 + 2x + 1 ) = 15
<=> ( x - 3 )( x2 - 6x + 9 - x2 - 3x - 9 ) + 9x2 + 18x + 9 = 15
<=> ( x - 3 ).(-9x) + 9x2 + 18x + 9 = 15
<=> -9x2 + 27x + 9x2 + 18x + 9 = 15
<=> 45x + 9 = 15
<=> 45x = 6
<=> x = 6/45 = 2/15
d) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 3
<=> x( x2 - 25 ) - ( x3 + 8 ) = 3
<=> x3 - 25x - x3 - 8 = 3
<=> -25x - 8 = 3
<=. -25x = 11
<=> x = -11/25
Bài 2.
a) A = 2( x3 + y3 ) - 3( x2 + y2 )
= 2( x + y )( x2 - xy + y2 ) - 3x2 - 3y2
= 2( x2 - xy + y2 ) - 3x2 - 3y2
= 2x2 - 2xy + 2y2 - 3x2 - 3y2
= -x2 - 2xy - y2
= -( x2 + 2xy + y2 )
= -( x + y )2
= -(1)2 = -1
b) B = x3 + y3 + 3xy
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 + 3xy
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
= ( x + y )3 - 3xy( x + y - 1 )
= 13 - 3xy( 1 - 1 )
= 1 - 3xy.0
= 1
Bài 1 :
\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)
\(\Leftrightarrow\)\(9x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{9}\)
Vậy \(x=\frac{16}{9}\)
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