Cho a, b, c là các số đôi một khác nhau t/m \(\left(a+b+c\right)^2=a^2+b^2+c^2\). Tính \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\)
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19 tháng 7 2016
a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\)
Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)
Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)
Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)
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 \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
17 tháng 4 2022
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)
-Ta có hằng đẳng thức: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(P=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}+2bc+2ca+2ab\)
\(=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}=\dfrac{2\left(b^3c^3+c^3a^3+a^3b^3\right)}{a^2b^2c^2}=\dfrac{2.\left(ab+bc+ca\right)\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=\dfrac{2.0.\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=0\)
17 tháng 4 2022
-C/m hằng đẳng thức trên:
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\left(đpcm\right)\)
3 tháng 8 2017
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
9 tháng 2 2019
\(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)
Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
Thay vào A ta được :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=0\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
\(\Leftrightarrow ab=-bc-ca\)
\(P=\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2+bc-bc-ca}=\frac{a^2}{\left(b-a\right)\left(c-a\right)}=\frac{-a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Tương tự rồi cộng tử lại xong đập bể ngoặc ra rồi dc tử bằng mẫu => P=1