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10 tháng 8 2016

a)a2+b2+c2+3=2(a+b+c)

=>a2+b2+c2+1+1+1-2a-2b-2c=0

=>(a2-2a+1)+(b2-2b+1)+(c2-2c+1)=0

=>(a-1)2+(b-1)2+(c-1)2=0

=>a-1=b-1=c-1=0 <=>a=b=c=1 

-->Đpcm

b)(a+b+c)2=3(ab+ac+bc)

=>a2+b2+c2+2ab+2ac+2bc -3ab-3ac-3bc=0 

=>a2+b2+c2-ab-ac-bc=0

=>2a2+2b2+2c2-2ab-2ac-2bc=0 

=>(a2- 2ab+b2)+(b2-2bc+c2) + (c2-2ca+a2) = 0

=>(a-b)2+(b-c)2+(c-a)2=0 

Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

c)a2+b2+c2=ab+bc+ca

=>2(a2+b2+c2)=2(ab+bc+ca)

=>2a2+2b2+c2=2ab+2bc+2ca

=>2a2+2b2+c2-2ab-2bc-2ca=0

=>a2+a2+b2+b2+c2+c2-2ab-2bc-2ca=0

=>(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ca+c2)=0

=>(a-b)2+(b-c)2+(a-c)2=0

Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

23 tháng 7 2019

a. \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)

\(\Leftrightarrow2a^2+2b^2=a^2+b^2-2ab\)

\(\Leftrightarrow a^2+b^2=-2ab\)

\(\Leftrightarrow a^2+2ab+b^2=0\)

\(\Leftrightarrow\left(a+b\right)^2=0\)

\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\) (đpcm)

b. \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

\(\left(a-1\right)^2;\left(b-1\right)^2;\left(c-1\right)^2\ge0\)

\(\Rightarrow\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)

\(\Leftrightarrow a-1=b-1=c-1=0\Leftrightarrow a=b=c=1\)

c. \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Tương tự câu b ta có a = b = c

31 tháng 12 2023

a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

=>\(\dfrac{a}{b}=\dfrac{5}{6}\)

b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

5 tháng 4 2018

Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

\(\Leftrightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2=0\)

\(\left\{{}\begin{matrix} -\left(a-b\right)^2\le0\\-\left(b-c\right)^2\le0\\-\left(c-a\right)^2\le0\end{matrix}\right.\Rightarrow-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\le0\)

Dấu ''= '' xảy ra \(\Leftrightarrow a=b=c\)

Vậy với a=b=c thì \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)

16 tháng 9 2017

\(https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7\)https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7
Ấn vào linh đấy ế

20 tháng 3 2021

a/ \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)

\(Tacó\dfrac{a^2+ab}{b^2+ab}=\dfrac{a\left(a+b\right)}{b\left(b+a\right)}=\dfrac{a}{b}\) (vì \(c^2=ab\) )

Vậy....

13 tháng 8 2015

=> a^2—2ab+b^2 +b^2-2bc+c^2+c^2-2ca+a^2-4a^2-4b^2-4c^2+4ab+4bc+4ca=0

=> —(2a^2+2^2+2c^2-2ab-2bc-2ca)=0

=>(a-b)^2+(b-c)^2+(c-a)^2=0

=>a=b;b=c;c=a

=>a=b=c