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Lời giải:
$a+b+c=0\Rightarrow a=-(b+c)\Rightarrow a^2=(b+c)^2$
$\Rightarrow b^2+c^2-a^2=b^2+c^2-(b+c)^2=-2bc$
$\Rightarrow \frac{1}{b^2+c^2-a^2}=\frac{1}{-2bc}=\frac{-1}{2bc}$
Hoàn toàn tương tự với các phân thức khác và cộng theo vế:
\(\text{VT}=\frac{-1}{2bc}+\frac{-1}{2ac}+\frac{-1}{2ab}=\frac{-(a+b+c)}{2abc}=\frac{-0}{2abc}=0\) (đpcm)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2-c^2=-2c^2-2bc-2ac-2ab\)
\(\Leftrightarrow a^2+b^2-c^2=-\left[2c.\left(c+b\right)+2a.\left(c+b\right)\right]\)
\(\Leftrightarrow a^2+b^2-c^2=-2.\left(a+c\right)\left(c+b\right)\)
Tương tự \(b^2+c^2-a^2=-2.\left(a+b\right)\left(a+c\right)\)
\(c^2+a^2-b^2=-2.\left(b+c\right)\left(b+a\right)\)
Đặt \(A=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=-\frac{1}{2}.\left[\frac{1}{\left(b+c\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(b+c\right)\left(a+b\right)}\right]\)
\(=-\frac{1}{2}.\frac{a+b+b+c+a+c}{\left(b+c\right).\left(a+c\right)\left(a+b\right)}=-\frac{1}{2}.\frac{2.\left(a+b+c\right)}{\left(b+c\right).\left(a+c\right).\left(a+b\right)}=0\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c+b+a}{abc}\right)\)
Mà a+b+c = 0 nên suy ra:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{0}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Ta có: (\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))\(^2\)= \(\frac{1}{a^2}\)+\(\frac{1}{b^2}\)+\(\frac{1}{c^2}\)+\(\frac{2}{abc}\)(\(\frac{a+b+c}{abc}\))
Mà
A+B+C= 0
nên: VT = VP (đpcm)
\(\frac{1}{a}+\frac{1}{c}=\frac{1}{a-b+c}+\frac{1}{b}\Leftrightarrow\frac{a+c}{ac}=\frac{a+c}{b\left(a-b+c\right)}\)
\(\Rightarrow\left[{}\begin{matrix}a+c=0\\ac=b\left(a-b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-c\\ac=b\left(a-b\right)+bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\ac-bc-b\left(a-b\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-c\\\left(c-b\right)\left(a-b\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\a=b\left(l\right)\\b=c\left(l\right)\end{matrix}\right.\) do \(a< b< c\) \(\Rightarrow a=-c\)
\(\Rightarrow\frac{1}{a^{2019}}-\frac{1}{b}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}}-\frac{1}{b}-\frac{1}{a^{2019}}=\frac{-1}{b}\)
\(\frac{1}{a^{2019}-b+c^{2019}}=\frac{1}{a^{2019}-b-c^{2019}}=\frac{-1}{b}\)
\(\Rightarrow\frac{1}{a^{2019}}-\frac{1}{b}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}-b+c^{2019}}\)
Bài 1:
a) Từ đkđb:
$x+y+z=0\Rightarrow x+y=-z; y+z=-x; z+x=-y$
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\Rightarrow xbc+yac+zab=0$
$a+b+c=0\Rightarrow a=-(b+c)\Rightarrow a^2=(b+c)^2$
$\Rightarrow a^2x=(b+c)^2x$.
Tương tự: $b^2y=(a+c)^2y; c^2z=(a+b)^2z$
Do đó:
$a^2x+b^2y+c^2z=(b+c)^2x+(a+c)^2y+(a+b)^2z=a^2(y+z)+b^2(z+x)+c^2(x+y)+2(xbc+yac+zab)$
$=a^2(-x)+b^2(-y)+c^2(-z)+2.0=-(a^2x+b^2y+c^2z)$
$\Rightarrow 2(a^2x+b^2y+c^2z=0$
$\Rightarrow a^2x+b^2y+c^2z=0$ (đpcm)
b)
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \frac{x+y+z}{2}=ax+by+cz\)
\(\Rightarrow \left\{\begin{matrix} ax=\frac{x+y+z}{2}-x=\frac{y+z-x}{2}\\ by=\frac{x+y+z}{2}-y=\frac{x+z-y}{2}\\ cz=\frac{x+y+z}{2}-z=\frac{x+y-z}{2}\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} a=\frac{y+z-x}{2x}\\ b=\frac{x+z-y}{2y}\\ c=\frac{x+y-z}{2z}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+1=\frac{y+z+x}{2x}\\ b+1=\frac{x+z+y}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=2\) (đpcm)
Bài 2:
Đặt $\frac{a_2}{a_1}=x; \frac{b_2}{b_1}=y; \frac{c_2}{c_1}=z$
Khi đó bài toán trở thành: Cho $x,y,z\neq 0$ thỏa mãn \(\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\\ x+y+z=1\end{matrix}\right.\)
CMR: $x^2+y^2+z^2=1$
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Thật vậy:
Ta có: \(\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\\ x+y+z=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} xy+yz+xz=0\\ x+y+z=1\end{matrix}\right.\)
Khi đó: $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=1^2-2.0=1$ (đpcm)
Vậy........
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow a+b+c=0\)
Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Anh/chị tham khảo ở đây nhé :
đặt x=a^2 + 2bc, y=b^2 + 2ac, z=c^2 + 2ab
=> x + y + z = (a + b + c)^2 <(=) 1
VT bpt : 1/x + 1/y + 1/z >(=) 3.căn3(1/xyz)...dùng cô-si cho 3 số nhé
mà x + y + z >(=) 3.căn3(xyz) <(=) 1
<=> 1/( 3.căn3 (xyz) >(=) 1 (ở đây là đổi chiều bđt)
<=> 1/ căn3 (xyz) >(=) 3
=> VT: 1/x + 1/y + 1/z >(=) 3.3 = 9
Mọi người giải nhanh giúp mk nhé. Cảm ơn rất nhiều !