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1) \(2x.\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2+x-6\right)-\left(x^2-4\right)\)
\(=-15x+10\)
b) \(2x.\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x.\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x^3-8\right)\)
\(=2x^3+4x^2+2x-x^3+3x^2-3x+1-x^3+8\)
\(=7x^2-x+9\)
c) \(\left(x-5\right)\left(x+5\right)\left(x+2\right)-\left(x+2\right)^3\)
\(=\left(x+2\right).\left[\left(x-5\right)\left(x+5\right)-\left(x+2\right)^2\right]\)
\(=\left(x+2\right).\left(x^2-25-x^2-4x-4\right)\)
\(=\left(x+2\right)\left(-4x-29\right)\)
\(=-4x^2-37x-58\)
d) \(\left(x-3\right)^3+\left(x-5\right)\left(x^2+5x+25\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-9x^2+27x-27+\left(x^3-125\right)-\left(x^3-1\right)\)
\(=x^3-9x^2+27x-151\)
e) \(\left(x-1\right)^3-\left(x-2\right)\left(x^2-2x+4\right)+3x^2+2x\)
\(=x^3-3x^2+3x-1-\left(x^3-8\right)+3x^2+2x\)
\(=5x+7\)
Nhẩm ấy, ko nháp âu
\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2-2x+3x-6\right)-\left(x^2-4x+4x-16\right)\)
\(=2x^2-14x-x^2+x-6-x^2+16\)
\(=-13x-10\)
\(2x\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(-2x^3+4x^2+2x-x^3+3x^2-3x+1-x^2+4\)
\(=-3x^3+6x^2-x+5\)
1) -3x( x + 2 )2 + ( x + 3 )( x - 1 )( x + 1 ) - ( 2x - 3 )2
= -3x( x2 + 4x + 4 ) + ( x + 3 )( x2 - 1 ) - ( 4x2 - 12x + 9 )
= -3x3 - 12x2 - 12x + x3 + 3x2 - x -3 - 4x2 + 12x - 9
= ( -3x3 + x3 ) + ( -12x2 + 3x2 - 4x2 ) + ( -12x - x + 12x ) + ( -3 - 9 )
= -2x3 - 13x2 - x - 12
2) ( x - 3 )( x + 3 )( x + 2 ) - ( x - 1 )( x2 - 3 ) - 5x( x + 4 )2 - ( x - 5 )2
= ( x2 - 9 )( x + 2 ) - ( x3 - x2 - 3x + 3 ) - 5x( x2 + 8x + 16 ) - ( x2 - 10x + 25 )
= x3 + 2x2 - 9x - 18 - x3 + x2 + 3x - 3 - 5x3 - 40x2 - 80x - x2 + 10x - 25
= ( x3 - x3 - 5x3 ) + ( 2x2 + x2 - 40x2 - x2 ) + ( -9x + 3x - 80x + 10x ) + ( -18 - 3 - 25 )
= -5x3 - 38x2 - 76x - 46
3) 2x( x - 4 )2 - ( x + 5 )( x - 2 )( x + 2 ) + 2( x + 5 )2 + ( x - 5 )2
= 2x( x2 - 8x + 16 ) - ( x + 5 )( x2 - 4 ) + 2( x2 + 10x + 25 ) + x2 - 10x + 25
= 2x3 - 16x2 + 32x - ( x3 + 5x2 - 4x - 20 ) + 2x2 + 20x + 50 + x2 - 10x + 25
= 2x3 - 16x2 + 32x - x3 - 5x2 + 4x + 20 + 2x2 + 20x + 50 + x2 - 10x + 25
= ( 2x3 - x3 ) + ( -16x2 - 5x2 + 2x2 + x2 ) + ( 32x + 4x + 20x - 10x ) + ( 20 + 50 + 25 )
= x3 - 18x2 + 46x + 95
a) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
= \(\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
= \(x^4+4x^2+4-x^4+16\)
= \(4x^2+20\)
b) \(\left(x+2y\right)^2-\left(x-2y\right)^2\)
= \(\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\)
= \(4y\cdot2x=8xy\)
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)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
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
D=(x2+x+1)(x2-x+1)(x4-x2+1)(x8-x4+1)
\(=\left(\left(x^2+1\right)^2-x^2\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(\left(x^4+1\right)^2-x^4\right)\left(x^8-x^4+1\right).\)
\(=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)=\left(x^8+1\right)-x^8=x^{16}+x^8 +1\)