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\(A=a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]+4abc\)
\(=a\left(b-c+a\right)\left(b-c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)+4abc\)
\(=\left(a+b-c\right)\left(ab-ac-a^2-bc+ab-b^2\right)+c\left(a^2-2ab+b^2-c^2+4ab\right)\)
\(=\left(a+b-c\right)\left[-c\left(a+b\right)-\left(a-b\right)^2\right]+c\left[\left(a+b\right)^2-c^2\right]\)
\(=\left(a+b-c\right)\left(-ca-cb-a^2+2ab-b^2+ac+cb+c^2\right)\)
\(=\left(a+b-c\right)\left(c^2-\left(a-b\right)^2\right)\)
\(=\left(a+b-c\right)\left(c+a-b\right)\left(a+b-c\right)\)
a(b-c)^2+b(a-c)^2+c(a-b)^2- a^3 -b^3 -c^3 +4abc
=a[(b-c)^2-a^2)]+ b[(a-c)^2-b^2)]+c[(a-b)^2-c^2)]+4abc
=a[(b-c)^2-a^2)]+ b[(a+c)^2-b^2)]+c[(a-b)^2-c^2)]
=a(b-c-a)(b-c+a)+b(a+c-b)(a+b+c)+c(a+c...
=[-a(b-c+a)+b(a+b+c)+c(a-b-c)](a+c-b)
em cu tiếp tục phân tích cái vế trong ngoặc vuông đuọc (a+b-c)(b+c-a) la d'c em nha
dap so la :(a+c-b)(a+b-c)(b+c-a)
tick nha !!!
Phân tích đa thức thành nhân tử \(a\left(b+c\right)^2+b\left(c+a\right)^2+c\left(a+b\right)^2-4abc\)
bn tick cho mik trước đi mik giải chi tiết ra cho
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)
\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)
\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)
\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)
1.a^3-7a-6
<=>x^3+2x^2-2x^2-4x-3x-6
<=>x^2-2x-3(x+2)=(x^2+x-3x-3)(x+2)
<=>[(x-3)(x+1)](x+2)
<=>(x-3)(x+1)(x+2)=0
<=>x-3=0 <=>x=3 hoặc x+1=0<=>x=-1 hoặc x+2=0<=>x=-2
2. a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc
=a(b^2+2bc+c^2)+b(c^2+2ca+a^2)+c(a^2+2ab+b^2)-4abc
=ab^2+2abc+ac^2+bc^2+2abc+ba^2+ca^2+2abc+b^2-4abc
=ab^2+bc^2+ca^2+cb^2+6abc-4abc
=ab^2+bc^2+ca^2+cb^2+2abc
=a^3+b^3+c^3+2abc
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
nhân ngược lại ra hay đặt ản thì tùy nhé =))
\(a,c\left(a+b\right)^2+b\left(c+a\right)^2+a\left(b+c\right)^2-4abc\)
\(=c\left(a+b\right)^2+bc^2+2abc+a^2b+ab^2+2abc+ac^2-4abc\)
\(=c\left(a+b\right)^2+\left(bc^2+ac^2\right)+\left(a^2b+ab^2\right)\)
\(=c\left(a+b\right)^2+c^2\left(a+b\right)+ab\left(a+b\right)\)
\(=\left(a+b\right)\left(ac+cb+c^2+ab\right)\)
\(=\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)