Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Chứng minh rằng a=b=c
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac-4a^2-4b^2-4c^2+4ab+4bc+4ac=0\)
\(\Leftrightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)(đpcm)
Chứng minh đẳng thức:
a) Cho \(2\left(a^2+b^2\right)=\left(a-b\right)^2.\) Chứng minh rằng a; b là 2 số đối nhau.
b) Cho \(a^2+b^2+c^2+3=2\left(a+b+c.\right)\) Chứng minh rằng a = b = c = 1
c) Cho \(\left(a+b+c\right)^2=3\left(ab+ac+bc\right).\) Chứng minh rằng a = b = c
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. \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì \(\left(a-1\right)^2;\left(b-1\right)^2;\left(c-1\right)^2\ge0\)
\(\Rightarrow\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
\(\Leftrightarrow a-1=b-1=c-1=0\Leftrightarrow a=b=c=1\)
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
Cho\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4.\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Chứng minh rằng a=b=c
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\forall a;b;c}\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\)
Vậy \(a=b=c\)
cho\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4.\left(a^2+b^2+c^2-ab-ac-bc\right)\)
chứng minh:a=b=c
Có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2-4a^2-4b^2-4c^2+4ab+4ac+4bc=0\)
\(\Rightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Rightarrow-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(a-c\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)
Cho: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Chứng minh rằng: a =b = c.
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
<=>\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca\)\(=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
<=>\(0=2a^2+2b^2+2c^2-2ab-2bc-2ca\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\)<=> a-b=b-c=c-a <=> a=b=c
vế phải= \(2\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)
=\(2\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]\)
=\(2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
=>\(\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]-2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow-1\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
Cho các số thực dương a, b, c. Chứng minh rằng:
\(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)^2}\ge\frac{54\left(abc\right)^3}{\left(a+b+c\right)^2\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}}\)
\(\Leftrightarrow\left(\Sigma a\right)^4\left(\Sigma a^4b^4\right)\left[\Sigma c^2\left(a^2+b^2\right)^2\right]\ge54^2\left(abc\right)^6\)
Giả sử \(c=\text{min}\left\{a,b,c\right\}\)và đặt \(a=c+u,b=c+v\) thì nhận được một BĐT hiển nhiên :P
Theo BĐT AM-GM ta có:
\(c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)\ge3\sqrt[3]{\left(abc\right)^2\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\right]^2}\)
\(\ge3\sqrt[3]{\left(abc\right)^264\left(abc\right)^4}=12\left(abc\right)^2\)
=> \(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(a^2+c^2\right)^2}\ge2\sqrt{3}abc\)
Cũng theo BĐT AM-GM \(\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4\ge3\sqrt[3]{\left(ab\right)^4\left(bc\right)^4\left(ca\right)^4}=3\left(abc\right)^2\sqrt[3]{\left(abc\right)^2}\)
=> \(\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}\ge\sqrt{3}\cdot abc\sqrt[3]{abc}\)và \(\left(a+b+c\right)^2\ge9\sqrt[3]{\left(abc\right)^2}\)
=> \(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)^2}\cdot\left(a+b+c\right)^2\cdot\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}\)
\(\ge2\sqrt{3}\left(abc\right)\cdot\sqrt{3}\left(abc\right)\sqrt[3]{abc}\cdot9\sqrt[3]{\left(abc\right)^2}\ge54\left(abc\right)^3\)
Dấu "=" xảy ra <=> a=b=c
\(\hept{\begin{cases}54&A,B,C^2&\end{cases}}\)\(\sqrt[54]{454}.A.B.C\)\(\sqrt{AB^4+BC^4+CA^4}\)\(\Rightarrow AB=CA=BC^4\)nên ta sẽ lại là 54abc3
vậy suy ra \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) ta =\(\notin54\) chả việc gì dài dòng cả
Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4.\left(a^2+b^2+c^2-ac-bc-ca\right)\). Chứng minh rằng : a = b = c
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ac-bc-ca\right)\)
⇔ \(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=0\)
⇔ \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\) \(\forall a,b,c\)
⇒ \(\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
⇔ \(a=b=c\)
⇒ \(ĐPCM\)
Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\cdot\left(a^2+b^2+c^2-ab-ac-bc\right)\). Chứng minh rằng a=b=c .
Giúp mik vs m.n @!
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\\ \Leftrightarrow a^2-2ab+b^2+b^2-2bc-c^2+c^2-2ac+a^2\\ =4a^2+4b^2+4c^2-4ab-4ac-4bc\\ \Leftrightarrow0=2a^2+2b^2+2c^2-2ab-2ac-2bc\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\\ \Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\Leftrightarrow\left\{\begin{matrix}\left(a-b\right)^2=0\Leftrightarrow a-b=0\Leftrightarrow a=b\\\left(a-c\right)^2=0\Leftrightarrow a-c=0\Leftrightarrow a=c\\\left(b-c\right)^2=0\Leftrightarrow b-c=0\Leftrightarrow b=c\end{matrix}\right.\)
Vậy a=b=c
Chứng minh rằng :
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
c) \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)\(=\left(a^3+b^3\right)+\left(a^3-b^3\right)=2a^3\Rightarrowđpcm\)
\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(=\left(a^3+b^3\right)\Rightarrowđpcm\)
\(c,\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\Rightarrowđpcm\)
a) (a+b)(a2-ab+b2)+(a-b)(a2+ab+b2)
= a3+b3+a3-b3 = 2a3
b) a3+b3
= (a+b)(a2-ab+b2)
= (a+b)(a2- 2ab+b2)+ab
= (a+b)(a2-b2)+ab
a. Biến đổi vế trái:
(a+b)(a2−ab+b2)+(a−b)(a2+ab+b2)=a3+b3+a3−b3=2a3
=>VT bằng VP (đpcm)
b. Biến đổi vế phải:
(a+b)[(a−b)2+ab]=(a+b)[a2−2ab+b2+ab]
=(a+b)(a2−ab+b2)=a3+b3
=>VP bằng VT (đpcm)
c. Biến đổi vế phải:
(ac+bd)2+(ad−bc)2=a2c2+2abcd+b2d2+a2d2−2abcd+b2c2
=a2c2+b2d2+a2d2+b2c2=c(a2+b2)+d2(a2+b2)=(a2+b2)(c2+d2)
=>VP bằng VT (đpcm)
