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( a - b + c )2
= [ ( a - b ) + c ]2
= ( a - b )2 + 2( a - b )c + c2
= a2 - 2ab + b2 + 2ac - 2bc + c2
= a2 + b2 + c2 - 2ab - 2bc + 2ca ( đpcm )
\(\left(a-b+c\right)^2\)
\(=\left(a-b+c\right).\left(a-b+c\right)\)
\(=a.\left(a-b+c\right)-b.\left(a-b+c\right)+c.\left(a-b+c\right)\)
\(=a^2-ab+ac-\left(ab-b^2+bc\right)+ac-bc+c^2\)
\(=a^2-ab+ac-ab+b^2-bc+ac-bc+c^2\)
\(=a^2-2ab+2ac+b^2-2bc+c^2\)
\(=a^2+b^2+c^2-2ab-2bc+2ac\)
\(\Rightarrow\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\left(đpcm\right).\)
(a+b+c)2=a2+b2+c2
=>2(ab+bc+ac)=0
=>ab+bc+ac=0
=> bc=-ab-ac
=>\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ac-ab+bc}\)=\(\frac{a^2}{\left(a-c\right)\left(a-b\right)}\)
Tuong tu => \(\frac{b^2}{b^2+2ac}=....\)
\(\frac{c^2}{c^2+2ab}=...\)
=> \(\frac{a^2}{a^2+2bc}+....\)=\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)+...
=\(\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
=1
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta lại có:
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab+bc-ca}+\frac{b^2}{b^2-ab-bc+ca}+\frac{c^2}{c^2+ab-bc-ca}\)
\(=\frac{a^2}{\left(b-a\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(c-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(=-\left(\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\frac{\left(a-b\right)\left(c-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Ai có thể giải thích cho mình đoạn a^2/(a^2-ab+bc-ca) đc ko mình cảm ơn
( a + b + c ) ^2 = a^2+b^2+c^2 + 2(ab+ac+bc)
=> ab = -ac-bc
bc= -ab-ac
ac= -ab-bc
a^2 + 2bc = a^2 + 2bc - ( ab + ac + ac)
= a^2 + bc - ab - ac
= ( a-c) ( a-b)
b^2 + 2ca = ( c-b) ( a-b)
c^2 + 2ab = (b-c) (a-c)
A= a^2/ ( a-c) (a-b) + b^2/ ( c-b) (a-b) + c^2/ ( b-c)(a-c)
rồi quy đồng là xong
\(\left(a+b+c\right)^2=[\left(a+b\right)+c]^2\)
\(=\left(a+b\right)^2+2.\left(a+b\right).c+c^2\)
\(=a^2+2ab+b^2+2ac+2bc+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\left(a+b+c\right)^2=a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)\)
\(=a^2+ab+ac+ab+b^2+bc+ac+bc+c^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc\)
Đặt A = a + b
Biến đổi vế trái ta có
:\(\left(A+c\right)^2=A^2+2Ac+c^2\)=\(\left(a+b\right)^2+2\left(a+b\right)c+c^2=a^2+b^2+2ab+2ac+2bc+c^2\)
Vậy vế trái bằng vế phải đẳng thức được chứng minh
Áp dụng BĐt Bunhiacopski dạng phân thức:
\(\text{Σ}_{cyc}\frac{a^2}{a^2+2bc}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)
\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" khi a = b = c
Bài làm:
Ta có: \(\left(a-b-c\right)^2\)
\(=\left[a-\left(b+c\right)\right]^2\)
\(=a^2-2a\left(b+c\right)+\left(b+c\right)^2\)
\(=a^2-2ab-2ac+b^2+2bc+c^2\)
\(=a^2+b^2+c^2-2ab+2bc-2ac\)
( a - b - c )2
= [ ( a - b ) - c ]2
= ( a - b )2 - 2( a - b )c + c2
= a2 - 2ab + b2 - 2ac + 2bc + c2
= a2 + b2 + c2 - 2ab + 2bc - 2ac ( đpcm )
BĐT cần chứng minh tương đương:
\(a^2+b^2+c^2\ge2ab-2bc+2ca\)
\(\Leftrightarrow a^2+b^2+c^2+2bc-2a\left(b+c\right)\ge0\)
\(\Leftrightarrow a^2+\left(b+c\right)^2-2a\left(b+c\right)\ge0\)
\(\Leftrightarrow\left(a-b-c\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng