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, bạn nào biết thì giúp mình sửa nhé
a, Theo bài ra ta có:
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
b, theo bài ra ta có:
\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)
\(=x^2\left(x-3\right)-2x\left(x-3\right)-3\left(x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x-3\right)\)
c,Theo bài ra ta có:
\(=x^3+5x^2+3x^2+15x+2x+10\)
\(=x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2+3x+2\right)\)
\(=\left(x+5\right)\left(x^2+x+2x+2\right)=\left(x+5\right)\left(x\left(x+1\right)+2\left(x+1\right)\right)\)
\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)
CHÚC BẠN HỌC TỐT...........
a) \(x^3-3x+2\)
= \(x^3-x^2+x^2-x-2x+2\)
= \(x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x-2\right)\)
= \(\left(x-1\right)\left(x^2+2x-x-2\right)\)
= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(x-1\right)\)
= \(\left(x-1\right)^2\left(x+2\right)\)
b) \(x^3-5x^2+3x+9\)
= \(x^3+x^2-6x^2-6x+9x+9\)
= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-6x+9\right)\)
= \(\left(x+1\right)\left(x-3\right)^2\)
c) \(x^3+8x^2+17x+10\)
= \(x^3+x^2+7x^2+7x+10x+10\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10\right)\)
= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)
= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
d) \(x^3-3x^2+6x+4\)
Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:
\(x^3+3x^2+6x+4\)
= \(x^3+x^2+2x^2+2x+4x+4\)
= \(x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+2x+4\right)\)
Học tốt nhé :))
a,
(x^2+x)^2+4x^2+4x-12
=x^4 + 2x^3 + 5x^2 + 4x -12
=(x-1)(x^3+3x^2+8x+12)
=(x-1)(x+2)(x^2+x+6)
b , 3x^2+6xy+3y^2-12
=3(x^2+2xy+y^2-4)
=3[(x+y)^2 -2^2]
=3(x+y+2)(x+y-2)
CMR :Tổng S=12+22+32+...+302 không phải là số chính phương.
Các bạn giúp mình nhé.....thanks các bạn 
Đầu tiên ta có tính chất của số chính phương là một số có dạng 3k+2 thì số đó ko thể nào là số chính phương nên tách tổng thành 10 nhóm, mỗi nhóm 3 số hạng
Ta có: \(S=\left(1^2+2^2+3^2\right)+\left(4^2+5^2+6^2\right)+...+\left(28^2+29^2+30^2\right)\)
\(=(3k_1+2)+\left(3k_2+2\right)+...+\left(3k_{10}+2\right)=3k+2\)
(trong đó \(k=k_1+k_2+...+k_{10}+6\)) từ đó ta có đpcm
Bài 1 :
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x^2-x\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x^2-x-6\right)\left(x+1\right)\)
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
câu a tìm mẫu thức chung, rồi đk là mtc khác 0
câu b rút dễ mà
câu c đập vô thôi
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
a, 4b2c2 - (b2+c2-a2)2
= (2bc)²-(b²+c²-a²)²
=(2bc+b²+c²-a²)(2bc-b²-c²+a²)
=[(b+c)²-a²][a²-(b-c)²]
=(b+c+a)(b+c-a)(a+b-c)(a-b+c).
b, 8x3-64 = 23.x3-43 = (2x)3-43
= (2x-4)[(2x)2+2.x.4+42] = (2x-4)(4x2+8x+16)
c, 8x3-27= 23.x3-33 = (2x)3-33
= (2x-3)[(2x)2+2.x.3+32] = (2x-3)(4x2+6x+9)