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1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)
27x6 + 125y6 = ( 3x2 )3 + ( 5y2 )3 = ( 3x2 + 5y2 )( 9x4 - 15x2y2 + 25y4 )
8a6 - 8b6 = ( 2a2 )3 - ( 2b2 )3 = ( 2a - 2b )( 4a2 + 4ab + 4b2 ) = 2( a - b )4( a2 + ab + b2 ) = 8( a - b )( a2 + ab + b2 )
x4 + 64y4 = x4 + 16x2y2 + 64y4 - 16x2y2
= ( x4 + 16x2y2 + 64y4 ) - 16x2y2
= ( x2 + 8y2 )2 - ( 4xy )2
= ( x2 + 8y2 - 4xy )( x2 + 8y2 + 4xy )
x4 + x3 + 2x2 + x + 1 = x4 + x3 + x2 + x2 + x + 1
= ( x4 + x3 + x2 ) + ( x2 + x + 1 )
= x2( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x2 + 1 )
\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)
\(8a^6-8b^6=8\left(a^6-b^6\right)=8\left(\left(a^3\right)^2-\left(b^3\right)^2\right)=8\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)
\(=\left(8y^2+x^2\right)^2-\left(4xy\right)^2=\left(8y^2+x^2+4xy\right)\left(8y^2+x^2-4xy\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
a) 16x2 - ( x2 + 4 )2
= ( 4x )2 - ( x2 + 4 )2
= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]
= ( -x2 + 4x - 4 )( x2 + 4x + 4 )
= [ -( x2 - 4x + 4 ) ]( x + 2 )2
= [ -( x - 2 )2 ]( x + 2 )2
b) ( x + y )3 + ( x - y )3
= [ ( x + y ) + ( x - y ) ][ ( x + y )2 - ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]
= 2x( 2x2 + 2y2 - x2 + y2
= 2x( x2 + 3y2 )
Áp dụng HĐT a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)
a, \(x^3+8y^3+27z^3-18xyz=x^3+\left(2y\right)^3+\left(3z\right)^3-3.x.2y.3z\)
\(=\left(x+2y+3z\right)\left[x^2+\left(2y\right)^2+\left(3z\right)^2-x.2y-2y.3z-3z.x\right]\)
\(=\left(x+2y+3z\right)\left(x^2+4y^2+9z^2-2xy-6yz-3xz\right)\)
các bài còn lại tương tự
1) 9(a+b)2-4(a-2b)2
=[3(a+b)]2-[2(a-2b)]2=....
2) 8x3+27y3=(2x)3+(3y)3=...
3) (a+b)3-c3=(a+b-c).[(a+b)2+(a+b)c+c2]
4) x3+3x2+3x+1=(x+1)3
\(a,\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)
\(b,27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
\(c,x^6-x^6\)
\(=0\)
\(d,10x-25-x^2\)
\(=-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)