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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)
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)
Ta có:a2(b3-c3)-a2(b-c)3
=a2(b3-3b2c+3bc2-c3+3b2c-3bc2)-a2(b-c)3
=a2(b-c)3+-a2(b-c)3-3a2bc(b-c)
=3abc(ab-ac)
Tương tự b2(c3-a3)-b2(c-a)3=3abc(bc-ab)
c2(a3-b3)-c2(a-b)3=3abc(ac-bc)
=>a2(b3-c3)-a2(b-c)3+b2(c3-a3)-b2(c-a)3+c2(a3-b3)-c2(a-b)3=3abc(ab-ac+bc-ab+ac-bc)=0
=>a2(b3-c3)+b2(c3-a3)+c2(a3-b3)=a2(b-c)3+b2(c-a)3+c2(a-b)3
Câu 4 :
Ta có : a+b+c=0
=> a+b=-c
Lại có : a3+b3=(a+b)3-3ab(a+b)
=> a3+b3+c3=(a+b)3-3ab(a+b)+c3
=-c3-3ab. (-c)+c3
=3abc
Vậy a3+b3+c3=3abc với a+b+c=0
a) Xét VP : \(4\left(p-b\right)\left(p-c\right)=4\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)\)
\(=\left(a+c-b\right)\left(a+b-c\right)=\left[a+\left(c-b\right)\right].\left[a-\left(c-b\right)\right]\)
\(=a^2-\left(b-c\right)^2=a^2-b^2-c^2+2bc=VT\)
b) Xét VT : \(p^2+\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2\)
\(=\left(\frac{a+b+c}{2}\right)^2+\left(\frac{a+b+c}{2}-a\right)^2+\left(\frac{a+b+c}{2}-b\right)^2+\left(\frac{a+b+c}{2}-c\right)^2\)
\(=\frac{\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2}{4}\)
\(=\frac{4\left(a^2+b^2+c^2\right)+2\left(ab+bc+ac+bc-ac-ab+ac-bc-ab+ab-bc-ac\right)}{4}\)
\(=\frac{4\left(a^2+b^2+c^2\right)}{4}=a^2+b^2+c^2=VP\)
a) Có a +b +c=0
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(ab+bc+ac\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left[a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right]\)\(\Rightarrow\)\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
1) Ta có a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (a2 - 2ac + c2) = 0
=> (a - b)2 + (b - c)2 + (a - c)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
a^2 + b^2 + c^2 = ab + bc + ca
<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 0
<=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
<=> a-b = 0 và b-c=0 và c-a=0
<=> a=b=c
a^2/b+c + b^2/a+c + c^2=a+b
= a(a/b+c) + b(b/a+c) + c(c/a+b)
= a(a/b+c + 1 - 1) + b(b/a+c + 1 - 1) + c(c/a+b + 1 - 1)
= a(a+b+c/b+c) - a + b(a+b+c/a+c) - b + c(a+b+c/a+b) - c
= (a+b+c)(a/b+c + b/a+c + c/a+b) - (A+b+c)
mà a/b+c + b/a+c + c/a+b = 1
= a+b+c - (a+b+c)
= 0
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac.\)\(\Rightarrow VT=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right).\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\left(đpcm\right)\)