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

=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)

=> \(\left(ab+bc+ac\right)^2=1\)

Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)

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

=> \(a^2b^2+b^2c^2+c^2a^2=1\)

Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

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

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

=> \(a^4+b^4+c^4=4-2=2\)

Ta có :

( a + b + c )² = a² + b² + c² + 2ab + 2 bc+ 2ac = 0

Mà a² + b ² + c² = 1 

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

=> ab + bc + ac = \(\frac{-1}{2}\)

=> ( ab + bc + ac ) ² = a²b² + a²c² + b²c ² + 2ab²c + 2ac²b + 2a²bc = \(\left(\frac{-1}{2}\right)^2\)=\(\frac{1}{4}\)

=> a²b² + a²c² + b²c² + 2abc ( a + b +c ) = \(\frac{1}{4}\)

mà a + b + c =  0 => 2abc ( a +b +c ) = 0

=> a²b² + b²c² + c²a² = \(\frac{1}{4}\)

Ta có : ( a² + b² + c² )² = a⁴ + b⁴ + c⁴ + 2 ( a²b² + b²c² + c²a² ) = 1

=> a⁴ +b⁴ + c⁴  + 2. \(\frac{1}{4}\) = 1

=> a⁴ + b⁴ + c⁴ = 1 - \(\frac{1}{2}\)

=> a⁴ + b⁴ + c⁴ = \(\frac{1}{2}\)

\(\Rightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{4}{2}=-2\)

\(\Leftrightarrow a^4+b^4+c^4=16-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

Mặt khác : \(\left(ab+bc+ac\right)^2=4\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=4\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=4\)

\(\Rightarrow a^4+b^4+c^4=16-2.4=8\)

a)Ta có 7x=2y

Suy ra:\(\dfrac{x}{\dfrac{1}{7}}\)=\(\dfrac{y}{\dfrac{1}{2}}\)

Và x-y=16

Áp dụng công thức của dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{\dfrac{1}{7}}\)=\(\dfrac{y}{\dfrac{1}{2}}\)=\(\dfrac{x-y}{\dfrac{1}{7}-\dfrac{1}{2}}\)=\(\dfrac{16}{\dfrac{-5}{14}}\)=\(\dfrac{-224}{5}\)

Từ \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{-224}{5}\)suy ra :x=\(\dfrac{-224}{5}\cdot\dfrac{1}{7}\)=\(-\dfrac{32}{5}\)

\(\dfrac{y}{\dfrac{1}{2}}=-\dfrac{224}{5}\)suy ra:y=\(-\dfrac{224}{5}\cdot\dfrac{1}{2}=-\dfrac{112}{5}\)

c)Ta có :\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

Mà a+2b-c=-20

Suy ra:\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{c}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có:

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{c}{4}=\dfrac{a+2b-c}{2+6-4}=-\dfrac{20}{4}=-5\)

Từ \(\dfrac{a}{2}=-5,suyra:a=-5\cdot2=-10\)

\(\dfrac{b}{3}=-5,suyra:b=-5\cdot3=-15\)

\(\dfrac{c}{4}=-5,suyra:c=-5\cdot4=-20\)

Vậy a=-10,b=-15,c=-20

bằng 1/2 bạn ơi

Ta có a + b + c = 0 

<=> (a + b + c)² = 0

<=> a² + b² + c² + 2(ab + bc + ca) = 0 

<=> ab + bc + ca = \(-\frac{1}{2}\)

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

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

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

Lại có a² + b² + c² = 1

=> (a² + b² + c²)² = 1

<= > a⁴ + b⁴ + c⁴ + 2[(ab)² + (bc)² + (ca)²] = 1 

<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)

<=> \(a^4+b^4+c^4=\frac{1}{2}\)

Từ a + b + c = 0 => ( a + b + c )² = 0 <=> a² + b² + c² + 2ab + 2bc + 2ca = 0

<=> ab + bc + ca = -1/2 => ( ab + bc + ca )² = 1/4

<=> a²b² + b²c² + c²a² + 2ab²c + 2bc²a + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc( a + b + c ) = 1/4

<=> a²b² + b²c² + c²a² = 1/4 ( vì a + b + c = 0 )

Từ a² + b² + c² = 1 => ( a² + b² + c² )² = 1 <=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 1

<=> a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 1 

<=> a⁴ + b⁴ + c⁴ + 1/2 = 1 <=> a⁴ + b⁴ + c⁴ = 1/2

Vậy A = 1/2

a: ĐKXĐ: x<>2; x<>-2

b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)

c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)