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\(a,\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)
\(b,27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
\(c,x^6-x^6\)
\(=0\)
\(d,10x-25-x^2\)
\(=-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
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 )
\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)
\(8a^6-8b^6=8\left(a^6-b^6\right)=8\left(\left(a^3\right)^2-\left(b^3\right)^2\right)=8\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)
\(=\left(8y^2+x^2\right)^2-\left(4xy\right)^2=\left(8y^2+x^2+4xy\right)\left(8y^2+x^2-4xy\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
c, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
\(=x^2\left(x+1\right)[x^2\left(x+1\right)-2x\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
d,
\(2x^3-x^2-1\)
\(=2x^3-2x^2+x^2-x+x-1\)
\(=2x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^2+x+1\right)\)
a) \(x^4-x^2y^2\) \(=x^2(x^2-y^2)\)
\(=x^2\left(x-y\right)\left(x+y\right)\)
b) \(6x^2y+3xy+9xy^2=3xy\left(2x+1+3y\right)\)
c) \(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\)
\(=\left(2x-5y\right)^2\)
d) \(x^3+6x^2+12x+8=x^3+3x^2.2+3x.2^2+2^3\)
\(=\left(x+2\right)^3\)
e) \(64x^3+y^3=\left(4x\right)^3+y^3\)
\(=\left(4x+y\right)\left(16x^2-4xy+y^2\right)\)
f) \(\dfrac{125}{216}a^3-8b^3=\left(\dfrac{5}{6}a\right)^3-\left(2b\right)^3\)
\(=\left(\dfrac{5}{6}a-2b\right)\left[\left(\dfrac{5}{6}a^2+\dfrac{5}{6}a.2b-\left(2b\right)^2\right)\right]\)
\(=\left(\dfrac{5}{6}a-2b\right)\left(\dfrac{25}{36}a^2+\dfrac{5}{6}ab-4b^2\right)\)
g) \(5xy-5x+10y-10=\left(5xy-5x\right)+\left(10y-10\right)\)
\(=5x\left(y-1\right)+10y\left(y-1\right)\)
\(=\left(y-1\right)\left(5x+10y\right)\)
h) \(x^2+2xy-z^2+y^2=\left(x^2+2xy+y^2\right)-z^2\)
\(=\left(x+y\right)^2-z^2\)
\(=\left(x+y-z\right)\left(x+y+z\right)\)
Hình như đề câu h bạn viết sai hay sao á,Mình nghĩ chắc là \(x^2\) thôi,vì \(x^3\) mình tính không ra :v Mong bạn xem lại đề có đúng như mình ns không
Học Tốt !!
a.\(x^2-64x=x\left(x-64\right)\)
b.\(24x^3-8=8\left(3x^3-1\right)\)
c.\(x^2-16y^2-3x+12y=\left(x^2-16y^2\right)-3\left(x-4y\right)\)\(=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)
k mình nha bn ^.^ thanks
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