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Mình tính thử a ,b ,c bằng nhau đó
Mình nghĩ là 0,037037037037037037
a) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ac\)
\(=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2ac\)
\(=b^2-2bc+2ac=b.\left(b-2c+2a\right)\)
b) \(x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\)
\(=\left(x-1\right)\left[x^2.\left(x+2\right)+x.\left(x+2\right)+6.\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
Pạn Khánh Châu ơi
Cái dòng thứ 2 đấy, dấu hiệu nhận biết là j vậy
Mà sao pạn phân tích hay vậy????
(a-b)3 + (b-c)3 + (c-a)3
=a3 - 3a2b + 3ab2- b3 + b3 - 3b2c + 3bc2- c3 + c3 - 3c2a + 3ca2- a3
=(-3a2b) + 3ab2 - 3b2c + 3bc2 - 3c2a +3ca2
=(-3a2b) + 3(ab2 - b2c + bc2 - c2a + ca2)
=(-3a2b) + 3[ab2 - b(bc - c2) - c(ca - a2)]
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Chúc bạn học tốt nha!!
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3ab\)
\(=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)
\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)
\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)
\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
a: \(a^3-a=a\left(a-1\right)\left(a+1\right)\)
Vì a;a-1;a+1 là ba số nguyên liên tiếp
nên \(a\left(a-1\right)\left(a+1\right)⋮3!\)
hay \(a^3-a⋮6\)
(a+b)3-(a-b)3=a3+3a2b+3ab2+b3-(a3-3a2b+3ab2-b3)
=a3+3a2b+3ab2+b3-a3+3a2b-3ab2+b3
=6a2b+2b3
Áp dụng hđt a3-b3=(a-b)(a2+ab+b2) ấy
\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
-- là giề
a(b^3-c^3) +b(c^3-a^3)+c(a^3-b^3)
=> a(b-c)(b^2+bc+c^2)+bc^3-ba^3+ca^3-cb^3
=>a(b-c)(b^2+bc+c^2)-(cb^3-bc^3)-(ba^3-ca^3)
=>a(b-c)(b^2+bc+c^2)-bc(b-c)(b+c)-a^3(b-c)
=>(b-c)(ab^2+abc+ac^2-cb^2-bc^2-a^3)
=>(b-c)(