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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: \(=\left(a+2\right)^3\)
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.
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 )
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: \(a^{2n+1}-b^{2n+1}=\left(a-b\right)\left(a^{2n}+a^{2n-1}.b+...+b^{2n-1}.a+b^{2n}\right)\)
\(=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^{\text{4}}\right)\)
1)\(4ax-6ay-8a^2\)
\(=2a\left(x-3y-4a\right)\)
2) \(3a^2x-12ax^2+18ax\)
\(=3ax\left(a-4x+6\right)\)
3) \(12ax^2y-18a^2xy^3-30axy^2\)
\(=6axy\left(2x-3ay^2-5y\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
\(a^3b-ab^3+a^2+2ab+b^2\)
\(=\left(a^3b-ab^3\right)+\left(a^2+2ab+b^2\right)\)
\(=ab\left(a^2-b^2\right)+\left(a+b\right)^2\)
\(=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\)
\(=\left(a+b\right)\left[ab\left(a-b\right)+\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)
bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
=( 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 ???