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Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)
A= bc(a+d)(b-c) +ac(b+d)(c-a) + ab(c+d)(a-b)
A= bc(ab+ bd -ac -dc ) + ac(bc+cd -ab-ad )+ab(ac+ad-bc-bd)
A=(ab²c + b²cd -abc² -bdc² ) + (abc² + adc² -a²bc -a²cd ) + (a²bc + a²bd - ab²c -ab²d)
A= (ab²c + cb²d -ab²c-ab²d) + (c²ab -abc² -bdc² +adc² ) + ( a²bd +a²bc -a²bc -a²cd)
A= a²(bd-cd) + b²(cd-ad) + c²(ad-bd)
A=a²d(b-c) + b²d(c-a) + c²d(a-b)
A=d(a²b-a²c + b²c-b²a +c²a-c²b)
A=d[b(a²-c²) + c(b²-a²) + a(c² - b²)]
Trả lời
P=(a+b+c)3-(a+b-c)3-(b+c-a)3-(c+a-b)3
Đặt a+b-c=x, b+c-a=y, c+a-b=z
=>(a+b+c)3-x3-y3-z3
Có x+y+z=a+b-c+b+c-a+c+a-b=a+b+c
=>(x+y+z)3-x3-y3-z3
=>[ (x+y)+z3 ]-x3-y3-z3
=>(x+y)3+z3+3z(x+y) (x+y+z)-x3-y3-z3
=>x3+y3+3xy(x+y)+z3+3z(x+y) (x+y+z)-x3-y3-z3
=>3(x+y) (xy+xz+yz+z2)
=>3(x+y)[x(y+z)+z(y+z)]
=3(x+y) (y+z) (x+z)
Áp dụng hằng đẳng thức trên ta có:
3(a+b-c+b+c-a) (b+c-a+c+a-b) (a+b-c+c+a-b)
=3.2b.2c.2a
=24abc
mk sẽ chỉ hướng để bạn làm bài
đầu tiên ta sẽ nhóm [ (a+b+c)3-(a+b+c)3 ] ở đây ta thấy có hằng đẳng thức
- [ (b+c-a)3 + ( c+a-b)3 ] đây cũng vậy
sau khi khai triển ta sẽ rút gọn sẽ có nhân tử là 2c
Phân tích đa thức sau thành nhân tử :
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=a^3\left(b-c\right)+b^3c-b^3a+c^3a-c^3b\\ \)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(bc\left(c-a\right)+b^2\left(c-a\right)-a\left(c^2-a^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-a\left(c+a\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-ac-a^2\right)\)
\(\left(b-c\right)\left(c-a\right)\left(b^2-a^2+c\left(b-a\right)\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(a+b+c\right)\)
Phân tích đa thức sau thành nhân tử
\(A=a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(A=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(A=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(A=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]\)
\(A=\left(b-c\right)\left(a-b\right)\left(a-c\right)\)
Auto cách khác:3
\(A=a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(c+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)