cho a,b,c đôi một khác nhau và thỏa điều kiện \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
rút gọn biểu thức:\(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
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5 tháng 6 2015
1) (a+b+c)^2=a^2+b^2+c^2 ab+bc+ca=0
<-->bc=−ac−ca -->a^2+2bc=a^2+bc−ca−ab
<--> a^2+2bc=(a−c)(a−b)
Tương tự với 2 phân số còn lại rồi quy đồng
2) Cộng hai vế của c^2+2(ab−ac−bc)=0 lần lượt với a^2;b^2 ta có:
a^2=c^2+2ab−2ac−2bc+a^2=(a−c)^2+2b(a−c) (1)
b^2=c^2+2ab−2ac−2bc+b^2=(b−c)^2+2a(b−c) (2)
Từ (1) và (2) -> $\frac{\text{a^2+(a−c)^2}}{\text{b^2+(b−c)^2}}=\frac{\text{(a−c)^2+2b(a−c)+(a−c)^2}}{\text{(b−c)^2+2a(b−c)+(b−c)^2}}=\frac{\text{2(a−c)^2+2b(a−c)}}{\text{2(b−c)^2+2a(b−c)}}=\frac{\text{2(a−c)(a−c+b)}}{\text{2(b−c)(b−c+a)}}=\frac{a-c}{b-c}$a^2+(a−c)^2b^2+(b−c)^2 =(a−c)^2+2b(a−c)+(a−c)^2(b−c)^2+2a(b−c)+(b−c)^2 =2(a−c)^2+2b(a−c)2(b−c)^2+2a(b−c) =2(a−c)(a−c+b)2(b−c)(b−c+a) =a−cb−c
KB
5 tháng 6 2015
1) (a+b+c)^2=a^2+b^2+c^2 ab+bc+ca=0
<-->bc=−ac−ca -->a^2+2bc=a^2+bc−ca−ab
<--> a^2+2bc=(a−c)(a−b)
Tương tự với 2 phân số còn lại rồi quy đồng
2) Cộng hai vế của c^2+2(ab−ac−bc)=0 lần lượt với a^2;b^2 ta có:
a^2=c^2+2ab−2ac−2bc+a^2=(a−c)^2+2b(a−c) (1)
b^2=c^2+2ab−2ac−2bc+b^2=(b−c)^2+2a(b−c) (2)
Từ (1) và (2) -> \(\frac{\text{a^2+(a−c)^2}}{\text{b^2+(b−c)^2}}=\frac{\text{(a−c)^2+2b(a−c)+(a−c)^2}}{\text{(b−c)^2+2a(b−c)+(b−c)^2}}=\frac{\text{2(a−c)^2+2b(a−c)}}{\text{2(b−c)^2+2a(b−c)}}=\frac{\text{2(a−c)(a−c+b)}}{\text{2(b−c)(b−c+a)}}=\frac{a-c}{b-c}\)
HP
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 \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
S
24 tháng 6 2018
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)
Thay (*) vào M ta được:
\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)
\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)
Vậy M = 0
NV
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 3 2020
cố tử thần ♡๖ۣۜŦεαм♡❤Ɠ长♡ღ
Chị ơi dùng bđt BCS , dấu = xảy ra P =1 như thế có gọi là giá trị của P=1 không nhỉ ?
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) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-ab-bc+ac}+\frac{c^2}{c^2-bc-ac+ab}\)
\(=\frac{a^2}{a\left(a-b\right)-c\left(a-b\right)}+\frac{b^2}{b\left(b-a\right)-c\left(b-a\right)}+\frac{c^2}{c\left(c-b\right)-a\left(c-b\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2a-b^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
19 tháng 8 2019
Tham khảo tại đây : Câu hỏi của Huỳnh Kim Bích Ngọc - Toán lớp 8 - Học toán với OnlineMath
Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\Rightarrow ab+ac+bc=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ac-ab\end{cases}}\)
Nên \(\frac{a^2}{a^2+2bc}=\frac{a^2+ab+bc+ac}{a^2+bc-ac-ab}=\frac{\left(a+c\right)\left(a+b\right)}{\left(a-c\right)\left(a-b\right)}\)
\(\frac{b^2}{b^2+2ac}=\frac{b^2+ab+bc+ac}{b^2+ac-ab-bc}=\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}\)
\(\frac{c^2}{b^2+2ab}=\frac{c^2+ab+ac+bc}{b^2+ab-ac-bc}=\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{\left(a+b\right)\left(a+c\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a+b\right)\left[\left(a+c\right)\left(b-c\right)+\left(b+c\right)\left(c-a\right)\right]+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a+b\right)\left(2bc-2ac\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{-2c\left(a+b\right)\left(a-b\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left[-2c\left(a+b\right)+\left(b+c\right)\left(c+a\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(-a^2+ab+c^2-bc\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Vậy \(P=1\)
cuối cùng P bằng 1 yên tâm mình tính rùi