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a, \(a^4+64=\left(a^2\right)^2+8^2=\left(a^2\right)^2+16a^2+8^2-16a^2\)
\(=\left(a^2+8\right)^2-\left(4a\right)^2=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
b, \(a^4+4b^2=\left(a^2\right)^2+\left(2b\right)^2=\left(a^2\right)^2+4a^2b+4b^2-4a^2b\)ĐK : b >= 0
\(=\left(a+2b\right)^2-\left(2a\sqrt{b}\right)^2=\left(a+2b-2a\sqrt{b}\right)\left(a+2b+2a\sqrt{b}\right)\)
p/s : mình nghĩ phần b bạn nên để đề là \(a^4+4b^4\)sẽ hợp lí hơn
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Trả lời:
1, x3 - x2 - 4
= x3 - x2 - 4 + 2x - 2x
= x3 - 2x2 + x2 - 4 + 2x - 2x
= ( x3 + x2 + 2x ) - ( 2x2 + 2x + 4 )
= x ( x2 + x + 2 ) - 2 ( x2 + x + 2 )
= ( x - 2 )( x2 + x + 2 )
2, x3 - 2x - 4
= x3 - 2x - 4 + 2x2 - 2x2
= x3 - 4x + 2x - 4 + 2x2 - 2x2
= ( x3 + 2x2 + 2x ) - ( 2x2 + 4x + 4 )
= x ( x2 + 2x + 2 ) - 2 ( x2 + 2x + 2 )
= ( x - 2 )( x2 + 2x + 2 )
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1, \(x^2+2x-3=x^2+3x-x-3=x\left(x-1\right)+3\left(x-1\right)=\left(x+3\right)\left(x-1\right)\)
2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
3, \(x^2-x-12=x^2-4x+3x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)
4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)
5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)
\(=4x\left(x+y\right)-9y\left(x+y\right)=\left(4x-9y\right)\left(x+y\right)\)
6, \(x^2-2x-4y^2-4y=x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)=\left(x-2y-2\right)\left(x+2y\right)\)
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a, \(4x^4+81=\left(2x^2\right)^2+9^2=\left(2x^2\right)^2+36x^2+9^2-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
b, \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
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\(a,x^3+x^2+4\)
\(x^3-x^2+2x+2x^2-2x+4\)
\(x\left(x^2-x+2\right)+2\left(x^2-x+2\right)\)
\(x^5+x+1\)
\(x^5-x^4+x^2+x^4-x^3+x+x^3-x^2+1\)
\(x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-3x^2+9\)
\(=\left(x^2\right)^2+2.x^2.3+3^2-9x^2\)
\(=\left(x^2+3\right)^2-\left(3x\right)^2\)
\(=\left(x^2-3x+3\right)\left(x^2+3x+3\right)\)
\(x^4+3x^2+4\)
\(=\left(x^2\right)^2+2.x^2.2+2^2-x^2\)
\(=\left(x^2+2\right)^2-x^2\)
\(=\left(x^2-x+2\right)\left(x^2+x+2\right)\)
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\(A=\left(5x^5+5x^4\right):5x^2-\left(2x^4-8x^2-6x+12\right):\left(2x-4\right)\)
Phép chia thứ nhất:
\(\left(5x^5+5x^4\right):5x^2=x^3+x^2\)
Phép chia thứ hai:
Vậy A = ( x^3 + x^2 ) - ( x^3 + 2x^2 - 3 ) = -x^2 + 3
Với x = -2 thì: A = -(-2)^2 + 3 = -4 + 3 = -1
B) bạn làm tương tự nhé
\(a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=\left(a^2+2b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+2b^2-2ab\right)\left(a^2+2b^2+2ab\right)\)
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)