(a2+b2+c2)2=196(a2+b2+c2)2=196
a4+b4+c4+2(a2b2+b2c2+c2a2)=196(1)a4+b4+c4+2(a2b2+b2c2+c2a2)=196(1)
ta lại có a+b+c)^2=0a2+b2+c2=−2(ab+bc+ca)=14a2+b2+c2=−2(ab+bc+ca)=14(ab+bc+ca)2=49(ab+bc+ca)2=49

a2b2+b2c2+c2a2+2abc(a+b+c)=49a2b2+b2c2+c2a2+2abc(a+b+c)=49
a2b2+b2c2+c2a2=49(2)a2b2+b2c2+c2a2=49(2)
Từ (1);(2)a4+b4+c4=196−49.2=98

bạn ghi tùm lum ko hiểu j hết  ghi lại được ko

Chia thành nhiều bước để tinh nha ban

Bước 1:a+b+c=0

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=0

1+2(ab+bc+ca)=0

2(ab+bc+ca)=-1

ab+bc+ca=-1/2

Bước 2 :(ab+bc+ca)^2=1/4

=a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=1/4

=a^2b^2+b^2c^2+c^2+a^2+2abc(a+b+c)=1/4

=a^2b^2+b^2c^2+c^2a^2+2abc.0

=a^2b^2+b^2c^2+c^2a^2

=>(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2=1/4

Bước 3:(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2

=a^4+b^4+c^4+2(a^2b^2+b62c^2+c^2a^2)=1

=a^4+b^4+c^4+2.1/4=1

=>a^4+b^4+c^4+1/2=1

=>a^4+b^4+c^4=1/2

Ta có: a+b+c=0

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

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

\(\Leftrightarrow2\left(ab+bc+ac\right)=0-1=-1\)

hay \(ab+bc+ac=-\dfrac{1}{2}\)

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

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

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

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

Ta có: \(M=a^4+b^4+c^4\)

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

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

\(\Leftrightarrow M=1^2-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy: \(M=\dfrac{1}{2}\)

Ta có : \(a+b+c=0\)

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

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)=1\) ( * )

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

Lại có : \(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\) ( suy ra từ * )

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

Vậy ...

\(a+b+c=0\)

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

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

\(\Leftrightarrow2ab+2bc+2ca=-14\)

\(\Leftrightarrow ab+bc+ca=-7\)

\(\Rightarrow\left(ab+bc+ca\right)^2=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).

\(a^2+b^2+c^2=14\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)

\(\Leftrightarrow a^4+b^4+c^4=98\)

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

\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)

Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)

\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=256\)  \(\Leftrightarrow a^4+b^4+c^4=98\)

Vậy ... 

http://olm.vn/hoi-dap/question/624161.html tối qua mình có giải cho 1 bạn bạn vào đây xem nha 

T I C K cho mình luôn nha mỉnh cảm ơn

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

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

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