Cho a > 2|b| . Chứng minh |a| < 2|a - b|
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy tích cho tui đi
vì câu này dễ mặc dù tui ko biết làm
Yên tâm khi bạn tích cho tui
Tui sẽ ko tích lại bạn đâu
THANKS
\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c\ge0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Dấu ''='' xảy ra <=> a = b = c = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a, a2+b2+c2+3=2(a+b+c)
a2+b2+c2+3-2a-2b-2c=0
(a2-2a+1)+(b2-2b+1)+(c2-2c+1)=0
(a-1)2+(b-1)2+(c-1)2=0
mà (a-1)2+(b-1)2+(c-1)2\(\ge\)0
=>\(\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
=> a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(a+b\right)^2=4ab\)<=> \(a^2+b^2+2ab=4ab\)
<=> \(a^2-2ab+b^2=0\)
<=> \(\left(a-b\right)^2=0\)=> a=b (đpcm)
b) Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
<=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)
<=> \(a^2y^2+b^2x^2-2axby=0\)
<=>\(\left(ay-bx\right)^2=0\)
<=>ay=bx(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow a=b=c=1\)
2/ \(\left(a+b+c\right)^2=3\left(a+b+c\right)\Leftrightarrow\left(a+b+c\right)\left(a+b+c-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a+b+c=3\end{cases}}\)
Bạn xem lại đề !