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LT
1
A
0
TT
1
30 tháng 7 2023
Từ đề bài ta có :
\(a+b+c=0< =>\left(a+b+c\right)^2=0< =>a^2+b^2+c^2+2ab+2ac+2bc=0\)
Mà \(a^2+b^2+c^2=1\) < = > 1 + 2 ( ab + ac + bc ) = 0
< = > 2 ( ab + ac + bc ) = -1
< = > ab + ac + bc = -1/2
\(< =>\left(ab+ac+bc\right)^2=\left(-\dfrac{1}{2}\right)^2< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)
\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)
\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2=\dfrac{1}{4}\)
Lại có từ \(a^2+b^2+c^2=1\)
\(< =>\left(a^2+b^2+c^2\right)^2=1< =>a^4+b^4+c^4+2\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]=1\)
\(< =>a^4+b^4+c^4+2.\dfrac{1}{4}=1< =>a^4+b^4+c^4+\dfrac{1}{2}=1< =>a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\left(đpcm\right)\)
TH
2
13 tháng 8 2017
a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) (1)
CẦn chứng minh:
2(a^4 + b^4 + c^4) = (a² + b² + c²)²
<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²)
<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²)
<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) )
<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1))
<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²)
<=> 8.(ab²c + bc²a + a²bc) = 0
<=> 8abc.(a + b + c) = 0
<=> 0 = 0 (đúng), Vì a + b + c = 0
=> Đpcm
EJ
13 tháng 8 2017
a + b + c = 0
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2.\left(ab+bc+ca\right)\left(1\right)\)
Cần phải chứng minh
2.(a4 + b4 + c4)=(a2+b2+c2)
\(\Leftrightarrow\) 2.(a4 - b4+c4)=a4+b4+c4+2.(a2b2+b2c2+c2a2)
\(\Leftrightarrow\)a4 +b4+c4=2.(a2b2+b2c2+c2a2)
\(\Leftrightarrow\) (a2 + b2 +c2 ) = 4(a2b2+b2c2 +c2a2)
\(\Leftrightarrow\) [ -2.(ab+bc+ca)2 ] = 4(a2b2+b2c2 +c2a2)
\(\Leftrightarrow\) 4(a2b2+b2c2 +c2a2)+8.(ab2c +bc2a+a2bc)=4.(a2b+b2c2+c2+a2
\(\Leftrightarrow\) 8(ab2c+bc2a+a2bc)=0
\(\Leftrightarrow\)8abc.(a+b+c)=0
\(\Leftrightarrow\) 0 =0 (đúng ) Vì a +b +c =0
=> ĐPCM
C
0
TT
1
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Ta có :
\(\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\left(1\right)\)
\(\Leftrightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))
\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)
\(\Rightarrow dpcm\)
8 tháng 3 2019
Từ a+b+c=6 \(\Rightarrow\)a+b=6-c
Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9
\(\Leftrightarrow\)ab=9-c(a+b)
Mà a+b=6-c (cmt)
\(\Rightarrow\)ab=9-c(6-c)
\(\Rightarrow\)ab=9-6c+c2
Ta có: (b-a)2\(\ge\)0 \(\forall\)b, c
\(\Rightarrow\)b2+a2-2ab\(\ge\)0
\(\Rightarrow\)(b+a)2-4ab\(\ge\)0
\(\Rightarrow\)(a+b)2\(\ge\)4ab
Mà a+b=6-c (cmt)
ab= 9-6c+c2 (cmt)
\(\Rightarrow\)(6-c)2\(\ge\)4(9-6c+c2)
\(\Rightarrow\)36+c2-12c\(\ge\)36-24c+4c2
\(\Rightarrow\)36+c2-12c-36+24c-4c2\(\ge\)0
\(\Rightarrow\)-3c2+12c\(\ge\)0
\(\Rightarrow\)3c2-12c\(\le\)0
\(\Rightarrow\)3c(c-4)\(\le\)0
\(\Rightarrow\)c(c-4)\(\le\)0
\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)
*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)
*
BN
11 tháng 7 2018
Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2=4\left[a^2b^2+b^2c^2+2abc\left(a+b+c+\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (a + b + c = 0)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (1)
Mà \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\) (2)
Từ (1) và (2) \(\Rightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\)
=> đpcm
S
11 tháng 7 2018
Ta có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (vì a + b + c = 0)
\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\) (1)
Lại có: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (2)
Thay (1) vào (2) ta được:
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\left(đpcm\right)\)
11 tháng 7 2018
Ta có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\) (1)
Cần chứng minh: \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^4+b^4+c^4\right)\)
\(\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)=a^4+b^4+c^4\)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2\) (Cộng hai vế cho 2(a2b2+b2c2+a2c2)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left[-2\left(ab+bc+ac\right)\right]^2\) (vì (1))
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+abc^2+a^2bc\right)\)
\(\Leftrightarrow8\left(ab^2c+abc^2+a^2bc\right)=0\)
<=> 8abc (a+b+c) = 0
<=> 0 = 0 (Vì a+b+c = 0 ) (luôn luôn đúng)
Vậy => đpcm
co a+b+c=0 =>b+c=-a
suy ra (b+c)2=(-a)2 hay b2+2bc+c2 =a2
hay b2+c2-a2 =-2bc
Suy ra (b2 + c2 - a2 )2 =( -2bc)2
<=> b4 +c4 +a4 +2b2c2 -2a2b2 -2a2c2 = 4b2c2
<=> a4+b4+c4 =2a2b2+2b2c2+2c2a2
<=> 2(a4+b4+c4) = a4+b4+c4+2a2b2+2b2c2+2c2a2
<=> a2+b2+c2 =2(a4+b4+c4) (dpcm)