ch/m rằng \(A=a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\) )luôn khác 0 nếu a khác b khác c
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13 tháng 7 2019
Ta có: \(\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{z\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{x\left(a-b\right)\left(a-c\right)}-\frac{1}{y\left(a-b\right)\left(b-c\right)}+\frac{1}{z\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{xz\left(a-c\right)}{yxz\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{xy\left(a-b\right)}{zxy\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(a-c\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)\(=\frac{yz\left(b-c\right)-xz\left[\left(b-c\right)+\left(a-b\right)\right]+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(b-c\right)-xz\left(a-b\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(y-x\right)-\left(a-b\right)x\left(z-y\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(c+a-b-b-c+a\right)-\left(a-b\right)x\left(a+b-c-c-a+b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(2a-2b\right)-\left(a-b\right)x\left(2b-2c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)2z\left(a-b\right)-\left(a-b\right)2x\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(2z-2x\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{2\left(z-x\right)}{xyz\left(a-c\right)}=\frac{2\left(a+b-c-b-c+a\right)}{xyz\left(a-c\right)}\)
\(=\frac{2\left(2a-2c\right)}{xyz\left(a-c\right)}=\frac{2.2\left(a-c\right)}{xyz\left(a-c\right)}=\frac{4}{xyz}\Rightarrowđpcm\)
10 tháng 9 2023
\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)
\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)
21 tháng 7 2016
2,
A=a4(b-c)+b4(c-a)+c4(a-b)
=a4(b-c)+b4[c-b)-(a-b)]+c4(a-b)
=a4(b-c)-b4(b-c)+c4(a-b)-b4(a-b)
=(a4-b4)(b-c)+(c4-b4)(a-b)
=(a-b)(b-c)(a+b)(a2+b2)-(a-b)(b-c)(b+c)(b2+c2)
=(a-b)(b-c)(a3+b3+a2b+ab2-b3-c3-b2c-bc2)
=(a-b)(b-c)(a2c+b2c+c3+abc+bc2+c2a-a3-ab2-ac2-a2b-abc-a2c)
=(a-b)(b-c)(c-a)(a2+b2+c2+ab+bc+ca)
=1/2(a-b)(b-c)(c-a)(2a2+2b2+2c2+2ab+2bc+2ca)
=1/2(a-b)(b-c)(c-a)[(a+b)2+(b+c)2+(c+a)2] khác 0
22 tháng 7 2016
Theo mình 4 dòng cuối bài giải của Nguyễn Thiều Công Thành phải có dấu "-" (âm) ở trước biểu thức
PN
22 tháng 9 2017
Trần Hữu Ngọc Minh bn tham khảo nha:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}=\frac{a+b+c}{"b+c"+"a+c"+"a+b"}=\frac{a+b+c}{2."a+b+c"}\)
Xét 2 trường hợp, ta có:
\(\cdot TH1:a+b+c=0\)thì \(\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}}\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-1+-1+-1=-3\)
Không phụ thuộc vào các giá trị a,b,c 1:
\(\cdot TH2:a+b+c\ne0\)thì \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2."a+b+c"}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=a+c\\2c=a+b\end{cases}}\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
Không phụ thuộc vào các giá trị a,b,c 2
Từ 1 và 2 \(\Rightarrow\)đpcm
16 tháng 11 2018
\(\frac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}=\frac{a^4}{\left[\left(a-b\right)\left(a+b\right)+c^2\right]\left[\left(a-c\right)\left(a+c\right)+b^2\right]}\)
\(\frac{a^4}{\left[-c\left(a-b\right)+c^2\right]\left[-b\left(a-c\right)+b^2\right]}=\frac{a^4}{4bc\left(b+c\right)^2}=\frac{a^4}{4a^2bc}\)
Tương tự với 2 phân thức còn lại, ta cũng có : \(\frac{b^4}{b^4-\left(c^2-a^2\right)^2}=\frac{b^4}{4ab^2c};\frac{c^4}{c^4-\left(a^2-b^2\right)^2}=\frac{c^4}{4abc^2}\)
\(VT=\frac{a^4}{4a^2bc}+\frac{b^4}{4ab^2c}+\frac{c^4}{4abc^2}=\frac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}=\frac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\)
\(VT=\frac{a^3+b^3+c^3}{4abc}\)
Mà \(a+b+c=0\) nên \(a^3+b^3+c^3=3abc\) ( tự cm )
\(\Rightarrow\)\(VT=\frac{3abc}{4abc}=\frac{3}{4}\) ( đpcm )
Chúc bạn học tốt ~
27 tháng 11 2021
1.
Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)
\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)
\(\RightarrowĐpcm\)
Ta có : \(A=a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=a^4\left[-\left(c-a\right)-\left(a-b\right)\right]+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=-a^4\left(c-a\right)+b^4\left(c-a\right)-a^4\left(a-b\right)+c^4\left(a-b\right)\)
\(=\left(c-a\right)\left(b^4-a^4\right)+\left(a-b\right)\left(c^4-a^4\right)\)
\(=\left(c-a\right)\left(b-a\right)\left(b+a\right)\left(b^2+a^2\right)+\left(a-b\right)\left(c-a\right)\left(c+a\right)\left(c^2+a^2\right)\)
\(=\left(c-a\right)\left(b-a\right)\left[\left(a+b\right)\left(a^2+b^2\right)-\left(c+a\right)\left(c^2+a^2\right)\right]\)
\(=\left(c-a\right)\left(b-a\right)\left[a^3+b^3+ab\left(a+b\right)-c^3-a^3-ac\left(a+c\right)\right]\)
\(=\left(c-a\right)\left(b-a\right)\left(b-c\right)\left(a^2+b^2+c^2+ab+bc+ac\right)\)
\(=\left(c-a\right)\left(b-a\right)\left(b-c\right)\left[\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{2}\right]\)
Đến đây bạn tự làm nhé :)