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12 tháng 7 2016

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 : \(p^2+\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2\)

\(=\left(\frac{a+b+c}{2}\right)^2+\left(\frac{a+b+c}{2}-a\right)^2+\left(\frac{a+b+c}{2}-b\right)^2+\left(\frac{a+b+c}{2}-c\right)^2\)

\(=\frac{\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2}{4}\)

\(=\frac{4\left(a^2+b^2+c^2\right)+2\left(ab+bc+ac+bc-ac-ab+ac-bc-ab+ab-bc-ac\right)}{4}\)

\(=\frac{4\left(a^2+b^2+c^2\right)}{4}=a^2+b^2+c^2=VP\)

17 tháng 1 2022
Ngu kkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
10 tháng 7 2019

#)Giải :

Ta có : \(a+b+c=2p\)

\(\Rightarrow b+c=2p-a\)

\(\Rightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)

\(\Rightarrow b^2+c^2+2bc=4p^2-4pa+a^2\)

\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

\(\Rightarrowđpcm\)

17 tháng 8 2020

a) Áp dụng Cauchy Schwars ta có:

\(M=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{\left(a+b+c\right)^2}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

17 tháng 8 2020

b) \(N=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\ge\frac{\left(1+2+3\right)^2}{a+b+c+3}=\frac{36}{6}=6\)

Dấu "=" xảy ra khi: x=y=1

23 tháng 12 2020

Ta có: a+b+c=0

nên a+b=-c

Ta có: \(a^2-b^2-c^2\)

\(=a^2-\left(b^2+c^2\right)\)

\(=a^2-\left[\left(b+c\right)^2-2bc\right]\)

\(=a^2-\left(b+c\right)^2+2bc\)

\(=\left(a-b-c\right)\left(a+b+c\right)+2bc\)

\(=2bc\)

Ta có: \(b^2-c^2-a^2\)

\(=b^2-\left(c^2+a^2\right)\)

\(=b^2-\left[\left(c+a\right)^2-2ca\right]\)

\(=b^2-\left(c+a\right)^2+2ca\)

\(=\left(b-c-a\right)\left(b+c+a\right)+2ca\)

\(=2ac\)

Ta có: \(c^2-a^2-b^2\)

\(=c^2-\left(a^2+b^2\right)\)

\(=c^2-\left[\left(a+b\right)^2-2ab\right]\)

\(=c^2-\left(a+b\right)^2+2ab\)

\(=\left(c-a-b\right)\left(c+a+b\right)+2ab\)

\(=2ab\)

Ta có: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)

\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

\(=\dfrac{a^3+b^3+c^3}{2abc}\)

Ta có: \(a^3+b^3+c^3\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-cb+c^2\right)-3ab\left(a+b\right)\)

\(=-3ab\left(a+b\right)\)

Thay \(a^3+b^3+c^3=-3ab\left(a+b\right)\) vào biểu thức \(=\dfrac{a^3+b^3+c^3}{2abc}\), ta được: 

\(M=\dfrac{-3ab\left(a+b\right)}{2abc}=\dfrac{-3\left(a+b\right)}{2c}\)

\(=\dfrac{-3\cdot\left(-c\right)}{2c}=\dfrac{3c}{2c}=\dfrac{3}{2}\)

Vậy: \(M=\dfrac{3}{2}\)

26 tháng 3 2018

Do a+b+c= 0

<=> a+b= -c 

=> (a+b)2= c2 

Tương tự: (c+a)2= b2, (c+b)2= a2   

Ta có: \(A=\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)

\(=\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(c+a\right)^2}+\frac{1}{a^2+b^2-\left(a+b\right)^2}\)

\(=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}\)

\(=\frac{a+b+c}{-2abc}=0\)