A/b=c/d cmr:
a) a^2+c^2/b^2+c^2=a/b
b) b^2 - a^2/a^2+c^2= b-a/a
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Y
0
7 tháng 7 2019
a) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(a^2+b^2+c^2+2ab+2ac+2bc-3ab-3ac-3bc=0\)
\(a^2+b^2+c^2-ab-ac-bc=0\)
\(2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
MC
1
TM
20 tháng 10 2016
(a-b)2+(b-c)2+(c-a)2=4(a2+b2+c2-ab-ac-bc)
<=>a2-2ab+b2+b2-2bc+c2+c2-2ac+a2=4a2+4b2+4c2-4ab-4ac-4bc
<=>2a2+2b2+2c2-2ab-2bc-2ac-4a2-4b2-4c2+4ab+4ac+4bc=0
<=>2ab+2ac+2bc-2a2-2b2-2c2=0
<=>-[(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ac+c2)]=0
<=>(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ac+c2)=0
<=>(a-b)2+(b-c)2+(c-a)2=0
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2}+\left(a-c\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)
<=>a-b=b-c=a-c
<=>a=b=c(đpcm)
HD
1
VM
25 tháng 10 2019
Ta có: \(\frac{\left(a^2+b^2\right)}{\left(c^2+d^2\right)}=\frac{ab}{cd}.\)
\(\Rightarrow\left(a^2+b^2\right).cd=ab.\left(c^2+d^2\right)\)
\(\Rightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Rightarrow a^2cd+b^2cd-abc^2-abd^2=0\)
\(\Rightarrow\left(a^2cd-abc^2\right)-\left(abd^2+b^2cd\right)=0\)
\(\Leftrightarrow ac.\left(ad-bc\right)-bd.\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right).\left(ac-bd\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\left(đpcm\right).\)
Chúc bạn học tốt!
22 tháng 10 2016
a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
\(\Rightarrow a=b=c\)
LT
0
LT
1
2 tháng 6 2018
Đặt \(A=\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\)
\(\Rightarrow A\ge\frac{a+b+c}{\left(b+c\right)^2+\left(c+a\right)^2+\left(a+b\right)^2}\)
\(\Rightarrow A\ge\frac{a+b+c}{b^2+2bc+c^2+c^2+2ac+a^2+a^2+2ab+b^2}\)
\(\Rightarrow A\ge\frac{a+b+c}{2\left(a^2+b^2+c^2\right)+2\left(ab+ac+bc\right)}\)
\(\Rightarrow A\ge\frac{a+b+c}{2\left[\left(a+b+c\right)^2-2\left(ab+ac+bc\right)\right]+2\left(ab+ac+bc\right)}\)
\(\Rightarrow A\ge\frac{a+b+c}{2\left(a+b+c\right)^2-2\left(ab+ac+bc\right)}\)
\(\Rightarrow A\ge1:\frac{2\left(a+b+c\right)^2-2\left(ab+ac+bc\right)}{a+b+c}\)
\(\Rightarrow A\ge1:\left[2\left(a+b+c\right)-\frac{2\left(ab+ac+bc\right)}{a+b+c}\right]\)
NT
0