(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

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

Ta có a + b + c = 0

=> a + b = -c

=> (a + b)² = (-c)²

=> a² + b² + 2ab = c²

=> a² + b² - c² = -2ab

=> (a² + b² - c²)² = (-2ab)²

=> a⁴ + b⁴ + c⁴ + 2a²b² - 2a²c² - 2b²c² = 4a²b²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2a²c²

Khi đó a² + b² + c² = 14

<=> (a² + b² + c²)² = 14²

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2a²c² = 196

=> a⁴ + b⁴ + c⁴ + a⁴ + b⁴ + c⁴ = 196 (Vì a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2a²c²)

=> 2(a⁴ + b⁴ + c⁴) = 196

=> a⁴ + b⁴ + c⁴ = 98

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

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

\(\Rightarrow1+2\left(ab+ac+bc\right)=0\)( vì \(a^2+b^2+c^2=1\))

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

\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)

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

Tới đây bạn phân tích nốt ra nhé :v

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

\(\Rightarrow a^2b^2+a^2c^2+b^2c^2=\frac{1}{4}\left(a+b+c=0\right)\)(*)

Mặt khác : \(a^2+b^2+c^2=\left(a^2+b^2+c^2\right)^2=1\)

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

\(\Rightarrow a^4+b^4+c^4+2\cdot\frac{1}{4}=1\)(theo *)

\(\Rightarrow a^4+b^4+c^4+\frac{1}{2}=1\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)

\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)

TA có \(\left(a+b+c\right)^2=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

=> \(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)

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

=> \(a^4+b^4+c^4=\frac{1}{2}\)

^_^

Ta có: a+b+c=0 <=> (a+b+c)²=0 <=> a²+b²+c²+ 2( ab+ac+bc)=0 <=> 2(ab+ac+bc)= -1 ( vì a²+b²+c²=1) <=> ab+ac+bc= -1/2 

=> (ab+ac+bc)²= 1/4 <=> a²b²+a²c²+b²c²+2abc(a+b+c)= 1/4 <=> 2(a²b²+a²c²+b²c²)= 1/2 ( vì a+b+c=0) (*)

Lại có: a²+b²+c²=1 <=> (a²+b²+c²)²=1 <=> a⁴+b⁴+c⁴+2(a²b²+a²c²+b²c²)=1 <=> a⁴+b⁴+c⁴= 1/2 ( vì (*))

Vậy,...

a² + b²  + c²=14

hay(a + b + c)² = 14

a⁴ + b⁴ + c⁴ =(a² + b² + c²).(a² + b² + c²)=(a+b+c)² . (a+b+c)² =14.14=196

k mk nha bạn kb nữa

1/ \(a+b+c=11\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=121\)

\(\Leftrightarrow ab+bc+ca=\frac{121-\left(a^2+b^2+c^2\right)}{2}=\frac{121-87}{2}=17\)

2/ \(a^3+b^3+a^2c+b^2c-abc\)

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

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

3/ \(x^4+3x^3y+3xy^3+y^4\)

\(=\left(\left(x+y\right)^2-2xy\right)^2-2x^2y^2+3xy\left(\left(x+y\right)^2-2xy\right)\)

\(=\left(9^2-2.4\right)^2-2.4^2+3.4.\left(9^2-2.4\right)=6173\)

bạn alibaba nguyễn có thể làm lại giúp mình được không ?