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24 tháng 5 2020

Dấu "=" xảy ra <=> a = b = c

24 tháng 5 2020

Áp dụng Bunhiacopxki dạng phân thức:

VT \(\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\) = 1

7 tháng 4 2017

Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\ge\frac{\left(a+b+c\right)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}\)=1

16 tháng 10 2020

Áp dụng bđt Cauchy-Schwarz dạng Engel ta có :

\(VT\ge\frac{\left(a+b+c\right)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)

=> đpcm

Dấu "=" xảy ra <=> a = b = c 

13 tháng 3 2017

a) đáp án A=1

b) B=0

c) C=1

3 tháng 1 2016

khó quá xin lỗi nha em  mới hok lớp 7

3 tháng 1 2016

Câu này lớp 7 tớ có làm. Cũng như cái mà gọi là áp dụng t/c dãy tỉ số bằng nhau và tỉ lệ thức. mình tính ra dc a, b. c rồi.

10 tháng 7 2016

a,b,c khác nhau đôi một nghĩa là từng cặp số khác nhau ,là:

+a khác b

+b khác c

+c khác a

\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0\)

Suy ra: \(ab==-\left(bc+ac\right)=-bc-ac\)

    \(bc=-\left(ab+ac\right)=-ab-ac\)

\(ac=-\left(ab+bc\right)=-ab-bc\)

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

Tương tự,ta cũng có: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\)

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

Vậy \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

10 tháng 7 2016

những câu còn lại tương tự,bn tự làm nhé
 

16 tháng 9 2020

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)

\(\Leftrightarrow bc+ca+ab=0\)

\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)

Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)

\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)

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

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

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

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

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

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

\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

3 tháng 12 2023

a, b, c chưa khác 0 bạn nhé

26 tháng 4 2017

bài 1 áp dụng bất đẳng thức Cô-si swatch ta có:

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}\)=1

dấu bằng xảy ra khi nào bạn tự tìm nh

19 tháng 1 2021

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(\frac{ab+bc+ac}{abc}=0\)

=> \(ab+bc+ac=0\)

=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)

a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

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

\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ca}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}-\frac{ca}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

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

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

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

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

19 tháng 1 2021

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

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

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

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

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

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

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

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

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

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

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