Ta có: a+b+c=0

nên \(\left(a+b+c\right)^2=0\)

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

\(\Leftrightarrow2ab+2ac+2bc=-1\)

\(\Leftrightarrow ab+ac+bc=\dfrac{-1}{2}\)

\(\Leftrightarrow\left(ab+ac+bc\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{1}{4}\)

Ta có: \(a^2+b^2+c^2=1\)

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

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1\)

\(\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\)

\(\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)

Vậy: \(a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}\)

1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

lại nhầm lần này đúng

(a+b+c)²=a²+b²+c²+2ac+2bc+2ab

=>0²=2+2(ac+bc+ab)

=>ac+bc+ab=2:2=-1

=>(-1)²=a²b²+b²c²+a²c²+2a²bc+2b²ac+2c²ab

(-1)²=a²b²+b²c²+a²c²+2abc(a+b+c)

=>1=a²b²+b²c²+a²c²+2abc.0

=>a²b²+b²c²+a²c²=1

(a²+b²+c²)²=a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²

(a²+b²+c²)²=a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)

2²=a⁴+b⁴+c⁴+2.1

4=a⁴+b⁴+c⁴+2

=>a⁴+b⁴+c⁴=2

trieu dang   làm sai đoạn cuối rồi

(a+b+c)²=a²+b²+c²+2ac+2bc+2ab

=>0²=2+2(ac+bc+ab)

=>ac+bc+ab=2:2=-1

=>(-1)²=a²b²+b²c²+a²c²+2a²bc+2b²ac+2c²ab

(-1)²=a²b²+b²c²+a²c²+2abc(a+b+c)

=>1=a²b²+b²c²+a²c²+2abc.0

=>a²b²+b²c²+a²c²=1

(a²+b²+c²)²=a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²

(a²+b²+c²)²=a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)

2²=a⁴+b⁴+c⁴+2.1

4=a⁴+b⁴+c⁴+2

=>a⁴+b⁴+c⁴=2

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

\(\Leftrightarrow2+2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=-1\Rightarrow\left(ab+bc+ca\right)^2=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=1\)

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

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc.0=1\)

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

Xét \(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(-1\right)=4\Leftrightarrow a^4+b^4+c^4=6\)

bạn ơi nó phải bằng 2 chứ

có a+b+c = 0 
=> a^2+b^2+c^2+2(ab+bc+ac) = 0
mà a^2+b^2+c^2 = 2
=> ab+bc+ac = -1
=> a^2b^2+b^2c^2+a^2c^2 + 2ab^2c+2a^2bc+2abc^2 = 1
=>a^2b^2+b^2c^2+a^2c^2 + 2abc(b+a+c) = 1
=>a^2b^2+b^2c^2+a^2c^2 = 1
Ta bình phong cái a^2+b^2+c^2 lên 
đk là
a^4+b^4+c^4 + 2a^2b^2+2a^2c^2+2b^2c^2=4
=> a^4+b^4+c^4 + 2(a^2b^2+a^2c^2+b^2c^2) = 4
mà ở trên là a^2b^2+b^2c^2+a^2c^2 = 1
=> a^4+b^4+c^4 +1 =4
a^4+b^4+c^4 = 3 

a) và b) là hai phần khác nhau nhé, ko phải là chung 1 phần đâu nha các bạn

để mk lm cho
 

a) Bình phương 2 vế của a+b+c=0 ta dược:

\(\left(a+b+c\right)^2=0^2\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

Mà \(a^2+b^2+c^2=2\)

\(\Rightarrow2\left(ab+bc+ac\right)=0-2=-2\Rightarrow ab+bc+ac=-\frac{2}{2}=-1\)

\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-1\right)^2=1\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1\)

Mà a+b+c=0

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1\)

Mặt khác,bình phương 2 vế của a²+b²+c²=2,ta được:

\(\left(a^2+b^2+c^2\right)^2=2^2\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\)

\(\Rightarrow a^4+b^4+c^4+2.1=4\Rightarrow a^4+b^4+c^4=4-2=2\)

b)tương tự,\(a^4+b^4+c^4=\frac{1}{2}\) nhé!

uk bai nay cung ko kho lam

[ a+ b + c] ^2 = a^2 + b^2 + c ^2 6 + 2[ ab + bc +ca] = 1 + 2 [ab +bc + ca] = 0

ab + bc + ca = - 1/2

[ a^ 2  + b ^ 2+ c^2]^2= a^ 4 + b^ 4 +c^4+2[ a^2 b^2+ b^2c^2 c^2a^2] = 1[ 1]

[ab + bc + ca] ^ 2 = a^2b^2 b^2c^2 c^2a^2 b^2 ac +c ^2ab + + a^bc = [a^2b^2+ b^2c^2 +c^2a^2] = 1/2

tu [1 ] va [2]  a^4 +b^4 + c^4 + 2[a^2b^2 + b^2c^2 + c^2+a^2] = a^4 + b^4 + c^4 + 1/2=1 hay a ^4 + b^ 4 + c^4 = 1/2

neu dung thi k mk nha. 

Bạn vào mục câu hỏi tương tự có rất nhiều.

a) Ta có: \(A\left(x\right)=ax^2+bx+c\)

Thay \(A\left(-1\right)\)  ta được:

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

\(=b-8-b=-8\)

b) \(\left\{{}\begin{matrix}A\left(0\right)=4\\A\left(1\right)=9\\A\left(2\right)=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b+c=9\\4a+2b+c=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\4a+2b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\2a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a=0\\b=5\end{matrix}\right.\)

c) 

Ta có: \(\left\{{}\begin{matrix}A\left(2\right)=4a+2b+c\\A\left(-1\right)=a-b+c\end{matrix}\right.\)

\(\Leftrightarrow A\left(2\right)+A\left(-1\right)=5a+b+2c=0\)

\(\Leftrightarrow A\left(2\right)=-A\left(-1\right)\)

\(\Leftrightarrow A\left(2\right)\times A\left(-1\right)=-\left[A\left(2\right)\right]^2\le0\)