Phân tích đt thành nt : 8 - 27x6y3
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\(ta\)\(có\)\(a^5+1=a^5+a^4-a^4-a^3+a^3+a^2-a^2+1\)
\(=a^4\left(a+1\right)-a^3\left(a+1\right)+a^2\left(a+1\right)+\left(-a+1\right)\left(a+1\right)\)
\(=\left(a+1\right)\left(a^4-a^3+a^2-a+1\right)\)
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a) Áp dụng hằng đằng thức hiệu của 2 bình phương ta có
\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
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1) Có 3 = (22 - 1)
=> BT = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 +1)
= (24 - 1)(24 + 1)(28 + 1)(216 +1)
= (28 - 1)(28 + 1)(216 +1)
= (216 - 1)(216 +1)
= 232 - 1
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\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
\(x^2-2x-3=x^2+x-3x-3=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)
a,x2-7x+12=(x-4)(x-3)
b,3x2+13x-10=(3x-2)(x+5)
c,x2-2x-3=(x-3)(x+1)
d,x2+3x-18=(x-3)(x+6)
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x2( x + 1 )2 + 2x2 + 2x - 8
= [ x( x + 1 ) ]2 + 2( x2 + x ) - 8
= ( x2 + x )2 + 2( x2 + x ) - 8 (*)
Đặt a = x2 + x
(*) = a2 + 2a - 8
= a2 - 2a + 4a - 8
= a( a - 2 ) + 4( a - 2 )
= ( a - 2 )( a + 4 )
= ( x2 + x - 2 )( x2 + x + 4 )
= ( x2 - x + 2x - 2 )( x2 + x + 4 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 4 )
= ( x - 1 )( x + 2 )( x2 + x + 4 )
ta co
\(x^2\left(x+1\right)^2+2x^2+2x-8\)
=\(\left(x\left(x+1\right)\right)^2+2x\left(x+1\right)+1-9\)
=\(\left(x^2+x+1\right)^2-9\)
=\(\left(x^2+x-8\right)\left(x^2+x+10\right)\)
=\(\left(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(\left(x+\frac{1}{2}\right)^2-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(x+\frac{1}{2}-\sqrt{\frac{33}{4}}\right)\left(x+\frac{1}{2}+\sqrt{\frac{33}{4}}\right)\left(x^2+x+10\right)\)
Ta có: \(8-27x^6y^3\)
\(=2^3-\left(3x^2y\right)^3\)
\(=\left(2-3x^2\right)\left(4+6x^2y+9x^4y^2\right)\)
\(8-27x^6y^3\)
\(=2^3-\left(3x^2y\right)^3\)
\(=\left(2-3x^2y\right)\left(4+6x^2y+9x^2y^2\right)\)