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Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
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 3:
a: \(=\left(x^3-1\right)\left(x^3-8\right)\)
\(=\left(1-1\right)\left(1-8\right)=0\)
b: \(=x^3-3x^2+3x-1-4x^3+4x+3\left(x^3-1\right)\)
\(=-3x^3-3x^2+7x-1+3x^3-3\)
\(=-3x^2+7x-4\)
\(=-3\cdot4-14-4=-30\)
Ta có:
\(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left[3x\left(2x+11\right)-5\left(2x+11\right)\right]-\left[2x\left(3x+7\right)+3\left(3x+7\right)\right]\)
\(=\left[\left(6x^2+33x\right)-\left(10x+55\right)\right]-\left[\left(6x^2+14x\right)+\left(9x+21\right)\right]\)
\(=\left[6x^2+23x-55\right]-\left[6x^2+23x+21\right]\)
\(=-55-21=-76\)
Vậy biểu thức A không phụ thuộc vào biến x, y.
B = (2x + 3).(4x^2 - 6x + 9) - 2.(4x^3 - 1)
B = (2x)^3 + 3^3 - 8x^3 + 2
B = 8x^3 - 8x^3 + 27 + 2
B = 0 + 27 + 2
B = 29
Vậy B không phụ thuộc vào x
a) ( x - 1 )3 - 4x( x + 1 )( x - 1 ) + 3( x - 1 )( x2 + x + 1 )
= x3 - 3x2 + 3x - 1 - 4x( x2 - 1 ) + 3( x3 - 13 )
= x3 - 3x2 + 3x - 1 - 4x3 + 4x + 3x3 - 3
= ( x3 - 4x3 + 3x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
b) ( x - 1 )( x - 2 )( 1 + x + x2 )( 4 + 2x + x2 )
= [ ( x - 1 )( 1 + x + x2 ) ][ ( x - 2 )( 4 + 2x + x2 ) ]
= [ ( x - 1 )( x2 + x + 1 ) ][ ( x - 2 )( x2 + 2x + 4 ) ]
= ( x3 - 13 )( x3 - 23 )
= ( x3 - 1 )( x3 - 8 )
= x6 - 9x3 + 8
c) ( x - 1 )3 + 3( x - 1 )( x2 + x + 1 ) - 4x( x + 1 )( x - 1 )
= x3 - 3x2 + 3x - 1 + 3( x3 - 13 ) - 4x( x2 - 1 )
= x3 - 3x2 + 3x - 1 + 3x3 - 3 - 4x3 + 4x
= ( x3 + 3x3 - 4x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
a,\(\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-3x^2+3x-1-4x\left(x^2-1\right)+3\left(x^3-1\right)\)
\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)
\(=-3x^2+7x-4\)
b,\(\left(x-1\right)\left(x-2\right)\left(1+x+x^2\right)\left(4+2x+x^2\right)\)
\(=\left[\left(x-1\right)\left(x^2+x+1\right)\right]\left[\left(x-2\right)\left(x^2+2x+4\right)\right]\)
\(=\left(x^3-1\right)\left(x^3-8\right)\)
\(=x^6-9x^3+8\)
c,\(\left(x-1\right)^3+3\left(x-1\right)\left(x^2+x+1\right)-4x\left(x+1\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1+3\left(x^3-1\right)-4\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1+3x^3-3-4x^3+4x\)
\(=-3x^2+7x-4\)