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=( 30a3+120a)-(18a2b+72b)
=30a(a2+1)-18b(a2+1)
=(30a-18b)(a2+1)
=6(5a-3b)(a2+1)
120 a để 30a làm thừa số chung thì phải còn 4 chứ nhỉ ???
Lời giải:
a)
$70a-84b-20ab-24b^2=2(35a-42b-10ab-12b^2)$
b)
$12y-9x^2+36-3x^2y$
$=(12y-3x^2y)+(36-9x^2)$
$=3y(4-x^2)+9(4-x^2)$
$=(3y+9)(4-x^2)=3(y+3)(2-x)(2+x)$
c)
$21bc^2-6c-3c^2+42b=3(7bc^2-2c-c^2+14b)$
d)
$30a^3-18a^2b-72b+120a$
$=(30a^3+120a)-(18a^2b+72b)$
$=30a(a^2+4)-18b(a^2+4)$
$=(a^2+4)(30a-18b)=6(a^2+4)(5a-3b)$
P/s: Cảm giác câu a,c bạn viết sai đề.
đây là hằng đẳng thức
\(a^3+6a^2+12a+8=a^3+3.2.a^2+3.2^2.a+2^3=\left(a+2\right)^3\)
x9 + 1
= (x3)3 + 13
= (x3 + 1)(x6 - x3 + 1)
= (x + 1)(x2 - x + 1)(x6 - x3 +1)
8a3 - 12a2 + 6a - 1
= (2a)3 - 3(2a)21 + 3 . 2a . 12 - 1
= (2a - 1)3
27a3 - 54a2b + 36ab2 - 8b3
= (3a)3 - 3(3a)22b + 3 . 3a . (2b)2 - (2b)3
= (3a - 2b)3
Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2\left[\left(ax^3+3ax^2\right)+\left(3ax+9a\right)\right]\)
\(=2a\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)
\(=2a\left(x+3\right)\left(x^2+3\right)\)
2ax3 + 6ax2 + 6ax + 18a
= 2a( x3 + 3x2 + 3x + 9 )
= 2a[ ( x3 + 3x2 ) + ( 3x + 9 ) ]
= 2a[ x2( x + 3 ) + 3( x + 3 ) ]
= 2a( x + 3 )( x2 + 3 )
k ) \(125x^3-1\)
\(=\left(5x\right)^3-1\)
\(=\left(5x-1\right)\left[\left(5x\right)^2+5x.1+1^2\right]\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m ) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right).\left[\left(x^2\right)^2+x^2.y+y^2\right]=\left(x^2-y\right).\left(x^4+x^2y+y^2\right)\)
n ) \(a^4-2a^2+1\)
\(=\left(a^2\right)^2-2.a^2.1+1^2=\left(a^2-1\right)^2\)
i ) \(a^3+6a^2+12a+8\)
\(=\left(a+2\right)^3\)
k) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
n) \(a^4-2a^2+1=\left(a^2-1\right)^2=\left(a^2-1\right)\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
i) \(a^3+6a^2+12a+8=\left(a+2\right)^2\)
Sửa ý đầu: \(\left(2a+3\right)x-\left(2a+3\right)y+2a+3\)
\(=\left(2a=3\right)\left(x-y+1\right)\)
\(\left(a-b\right)c+\left(b-a\right)y-a+b\)
\(=\left(a-b\right)c-\left(a-b\right)y-\left(a-b\right)\)
\(=\left(a-b\right)\left(c-y-1\right)\)
\(81a^2+18a+1\)
\(=\left(9a+1\right)^2\)
\(a^3-1\)
\(=\left(a-1\right)\left(a^2+a+1\right)\)
\(a^5-b^5\)
Áp dụng công thức: \(a^{2n+1}-b^{2n+1}=\left(a-b\right)\left(a^{2n}+a^{2n-1}.b+...+b^{2n-1}.a+b^{2n}\right)\)
\(=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^{\text{4}}\right)\)
a) \(6a^2b^2c-4ab^2c^2+12a^2bc^2\)
\(=2abc\left(3ab-2bc+6ac\right)\)
b)\(x^2\left(x-y\right)-y\left(y-x\right)\)
\(=x^2\left(x-y\right)+y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+y\right)\)
b. \(a^3+6a^2+12a+8\\ =a^3+3.a^2.2+3.2^2.a+2^3\\ =\left(a+2\right)^3\)
a: \(=30a\left(a^2+4\right)-18b\left(a^2+4\right)\)
\(=6\left(a^2+4\right)\left(5a-3b\right)\)
b: \(=\left(a+2\right)^3\)