Cho a+b+c+d= 0
CMR : \(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
cmr:
A)\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
B)\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
Chứng minh rằng nếu a + b + c + d = 0 thì
a)\(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
b)\(\left(b+d\right)\left(ac-bd\right)=\left(b+c\right)\left(cd-bc\right)\)
cmr:
A)\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
B)\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
Cho a+b+c=0
a) Chứng minh rằng: \(a^3+a^2c-abc+b^2c+b^3=0\)
b) Áp dụng phân tích thành nhân tử đa thức: \(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)
Cho : a + b + c + d = 0
Chứng minh rằng \(a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
Cho a, b, c > 0. CMR (dùng BĐT Schur) :
4\(\left(a+b+c\right)\)(ab + bc + ca) ≤ \(\left(a+b+c^{ }\right)^3\) + 9abc
CMR: \(\left(a+b+c+d\right)^2\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\\ \)
cho a,b,c>0 tm abc=1. cmr \(\dfrac{1}{a^3\left(b+c\right)}\) + \(\dfrac{1}{b^3\left(c+a\right)}\) +\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
