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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)]
a) \(5a^2-5ax-7a+7x\)
\(=5a\left(a-x\right)-7\left(a-x\right)\)
\(=\left(5a-7\right)\left(a-x\right)\)
c) \(x^2-\left(a+b\right).x+ab\)
\(=x^2-ax-bx+ab\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
Ta có:
\(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)
\(=bc\left(a+d\right)\left[\left(b-a\right)+\left(a-c\right)\right]-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\)\(\left(a-b\right)\)
\(=bc\left(a+d\right)\left(a-b\right)+bc\left(a+d\right)\left(a-c\right)-ac\left(b+d\right)\left(a-c\right)\)\(+ab\left(c+d\right)\left(a-b\right)\)
\(=b\left(a-b\right)\left[a\left(c+d\right)-c\left(a+d\right)\right]+c\left(a-c\right)\left[b\left(a+d\right)-a\left(b+d\right)\right]\)
\(=b\left(a-b\right).d\left(a-c\right)+c\left(a-c\right).d\left(b-a\right)\)
\(=d\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)
\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)
\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)
\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)
\(\Rightarrow B=x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)
\(x+y=a+b-2c+b+c-2a=2b-a-c\)
\(y+z=b+c-2a+c+a-2b=2c-a-b\)
\(z+x=c+a-2b+a+b-2c=2a-b-c\)
Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)
Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)
a:
\(a^3+a^2c-abc+b^2c+b^3\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)=0\)(vì a+b=c=0)
câu b bn xem ở link này nha!
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