cho A=\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
rut gon A biet a+b+c=0
K
Khách
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17 tháng 2 2018
Vì a+b+c=0. Suy ra
* a+b=-c
=> (a+b)2=c2
=> a2+b2+2ab=c2
=>a2+b2-c2=-2ab
tương tự ta đc a2+c2-b2=-2ac và c2+b2-a2=-2bc
Ta có
A=\(\dfrac{1}{a^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=>\(A=\dfrac{-1}{2bc}-\dfrac{1}{2ac}-\dfrac{1}{2ab}\)
=>A=\(\dfrac{-a}{2abc}-\dfrac{b}{2abc}-\dfrac{c}{2abc}\)
=>A=\(\dfrac{-a-b-c}{2abc}=\dfrac{-\left(a+b+c\right)}{2abc}\)
=>\(\dfrac{0}{2abc}=0\) (vì a+b+c=0)
vậy A=0
6 tháng 4 2018
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
3 tháng 1 2019
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 1 2019
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
17 tháng 11 2021
\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)
Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2018
Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Đầu tiên ta có hẳng đẳng thức:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow0=a^3+b^3+c^3+3\left(-c\right)\left(-b\right)\left(-a\right)\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\Rightarrow a^3+b^3+c^3=3abc\)
Biến đổi mẫu thức:
\(a^2-b^2-c^2=\left(a+b\right)\left(a-b\right)-c^2=-c\left(a-b\right)-c^2=-c\left(a-b+c\right)=2bc\)
Tương tự: \(b^2-c^2-a^2=2ac;\) \(c^2-a^2-b^2=2ab\)
\(\Rightarrow A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
H
1 tháng 4 2021
Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.
a) \(\dfrac{a}{1+b^2}=\dfrac{a\left(1+b^2\right)}{1+b^2}-\dfrac{ab^2}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\)
(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)
\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)
b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)
c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
30 tháng 12 2020
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
30 tháng 12 2020
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-c^2=-2ac\\b^2+c^2-a^2=-2bc\\\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{1}{-2ab}+\dfrac{1}{-2ac}+\dfrac{1}{-2bc}=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)