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a) a3+b3+a2c+b2c-abc
= (a+b)(a2-ab+b2)+c(a2+b2)-abc
=(a+b) [ (a+b)2-3ab]+c.[(a+b)2-2ab]-abc
=(a+b)(a+b)2-3ab(a+b)+c(a+b)2-3abc
=(a+b)2(a+b+c)-3ab(a+b+c)
=(a+b)2.0-3ab.0
=0
b) ax+ay+2x+2y+4
=a(x+y)+2(x+y)+4
=(x+y)(a+2)+4
=(a-2)(a+2)+4
=a2-4+4
=a2
c) A=1+x+x2+...+x49=>Ax=x+x2+x3+...+x50
- A=1+x+x2+...+x49
---> Ax-A=x50-1
d)(a+b)(a+c)+(c+a)(c+b)
=a2+ac+ab+bc+c2+bc+ac+ab
=a2+c2+2ac+2ab+2bc
=2b2+2bc+2ac+2ab
=2b(b+c)+2a(b+c)
=2b(b+c)(b+a)
Hai BĐT đều có dấu "=" xảy ra
a/ \(\Leftrightarrow x^7-x^4y^3+y^7-x^3y^4\ge0\)
\(\Leftrightarrow x^4\left(x^3-y^3\right)-y^4\left(x^3-y^3\right)\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x^3-y^3\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+y^2\right)\left(x^2+xy+y^2\right)\left(x-y\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(x=y\)
b/ Áp dụng câu a:
\(VT\le\sum\frac{a^2b^2}{a^3b^3\left(a+b\right)+a^2b^2}=\sum\frac{1}{ab\left(a+b\right)+1}=\sum\frac{abc}{ab\left(a+b\right)+abc}=\sum\frac{c}{a+b+c}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
cau 1 ne:
a^2 + b^2 + c^2 + 3
theo bat dang thuc cosi ban se co
a^2 + a + 1 >= 3a
b^2 + b + 1 >= 3b
c^2 + c + 1 >= 3c
cong 3 ve bat dang thuc lai voi nhau ban se co
a^2 + b^2 + c^2 + (a + b + c) + 3>= 3(a + b + c)
=> a^2 + b^2 + c^2 + 3 >= 2(a + b + c)
dau = xay ra <=> a= b= c = 1
ma theo de bai ta lai co a^2 + b^2 + c^2 + 3 = 2(a + b + c)
=> a = b = c = 1 (dpcm)
b) (a - b)^2 + (b-c)^2 + (c - a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
hay (a + b - 2b)^2 + (b + c - 2c)^2 + (c + a - 2a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
dat. a + b = A
b + c = B
c + a = C
=> ban se co:
(A - 2b)^2 + (B - 2c)^2 + (C - 2a)^2 = (A - 2c)^2 + (B - 2a)^2 + (C - 2b)^2
tu day ban nhan pha ra roi rut gon 2 ve cho nhau ban se co
Ab + Bc + Ca = Ac + Ba + Cb
hay (a + b)b + (b + c)c + (c + a)a = (a + b)c + (b + c)a + (c + a)b
hay ab + b^2 + bc + c^2 + ac + a^2 = 2ab + 2bc + 2ac
hay a^2 + b^2 + c^2 - ab - bc - ac = 0
hay 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
hay (a-b)^2 + (b-c)^2 +(c - a)^2 = 0
dau = xay ra <=> a = b = c (dpcm)
c) a^3 + b^3 + c^3 + d^3 = (a + b)(a^2 -ab +b^2) + (c+d)(c^2 - cd + d^2) (**)
ban nhan thay a + b + c + d = 0
=> a + b = - c - d
thay vao pt (**) ban se co
-(c + d)(a^2 - ab + b^2) + (c + d)(c^2 - cd + d^2)
(c + d)(c^2 - cd + d^2 -a^2 + ab - b^2)
hay (c + d)(ab - cd + (c^2 + d^2 - a^2 - b^2)) (***)
ban co a + b = - c - d
hay (a + b)^2 = (c + d)^2
hay a^2 + b^2 + 2ab = c^2 + d^2 + 2cd
hay c^2 + d^2 - a^2 - b^2 = 2ab - 2cd
thay vao pt (***) ban se co
(c + d)(ab - cd + 2ab - 2cd)
hay (c +d)(3ab - 3cd) = 3(c+d)(ab - cd) (dpcm)
\(VT=\left(a+b\right)\left(a+c\right)+\left(a+c\right)\left(c+b\right)=\left(a+c\right)\left(a+b+c+b\right)=\left(a+c\right)\left(a+c+2b\right)\)
\(=\left(a+c\right)^2+2b\left(a+c\right)=a^2+2ac+c^2+2ab+2bc=2b^2+2ac+2ab+2bc\)
\(VP=2\left(a+b\right)\left(b+c\right)=2ab+2ac+2b^2+2bc\)
\(\Leftrightarrow VT=VP\left(đpcm\right)\)