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Rút gọn biểu thức:
a/ (x2+2)2 - (x+2)(x-2)(x2+4)
= (x2+2)2- (x2-22)(x2+4)
= (x2+2)2- (x2-4)(x2+4)
=(x2+2)2- (x4- 42)
= x4+4x2+4-x4+16
= 4x2+20
=4(x2+5)
b/ (x+2y)2-(x-2y)2
= x2+4xy+4y2-x2+4xy-y2
= 8xy
A. \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^4+4x^2+4\right)-\left[\left(x^2-4\right)\left(x^2+4\right)\right]\)
\(=\left(x^4+4x^2+4\right)-\left(x^4+4x^2-4x^2-16\right)\)
\(=x^4+4x^2+4-x^4-4x^2+4x^2+16\)
\(=4x^2+20\)
B. \(\left(x+2y\right)^2-\left(x-2y\right)^2\)
\(=\left(x^2+4xy+4y^2\right)-\left(x^2-4xy+4y^2\right)\)
\(=x^2+4xy+4y^2-x^2+4xy-4y^2\)
\(=6xy\)
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)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Mình làm thử nha:
a/ \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right).\left(x-1\right)\)
\(=\left(x+1\right)\left(x+1\right)-\left(x-1\right)\left(x-1\right)-\left(3x+3\right).\left(x-1\right)\)
\(=\left[\left(x+1\right)\left(x+1\right)-\left(x-1\right)\left(x-1\right)\right]-4x+\left(-3\right)\)
Từ đó làm tiếp
b/ \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)^4+17\)
\(=\left(5x+10\right)\left(x-2\right)-\left(3-4x\right)^4+17\)
\(=6x+\left(-20\right)-\left(81-256x\right)+17\)
Làm nốt nha
\(A=x^2-2x+1-x^2+4=5-2x\)
\(B=27x^3+8-x^2+9=27x^3-x^2+17\)
\(C=3x^2y-6xy^2-2x\left(x^2-2x^2y+x^2y^2\right)=3x^2y-6xy^2-2x^3+4x^3y-2x^3y^2\)
Em chỉ cần nhớ hằng đẳng thức và áp dụng là biến đổi được ^^
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
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) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
1. ( 2x + y )( 4x2 - 2xy + y2 ) - 8x3 - y3 - 16
= [ ( 2x )3 + y3 ] - 8x3 - y3 - 16
= 8x3 + y3 - 8x3 - y3 - 16
= -16 ( đpcm )
2. ( 3x + 2y )2 + ( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 9x2 + 12xy + 4y2 ) - 18x2 - 8y2 + 3
= 18x2 + 24xy + 8y2 - 18x2 - 8y2 + 3
= 24xy + 3 ( có phụ thuộc vào biến )
3. ( -x - 3 )3 + ( x + 9 )( x2 + 27 ) + 19
= -x3 - 9x2 - 27x - 27 + x3 + 9x2 + 27x + 243 + 19
= -27 + 243 + 19 = 235 ( đpcm )
4. ( x - 2 )3 - x( x + 1 )( x - 1 ) + 13( x - 4 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 13x - 52
= x3 - 6x2 + 12x - 8 - x3 + x + 13x - 52
= -6x2 + 26x - 60 ( có phụ thuộc vào biến )
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) \(\left(x+2y\right)^2-\left(x-2y\right)^2\)
= \(\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\)
= \(4y\cdot2x=8xy\)