giúp em với ạ 8:45 em đi học thêm rồi

a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2((ab)^2+(bc)^2+(ca)^2) = 1 - 2((ab)^2+(bc)^2+(ca)^2) (*) 
Do a+b+c = 0 => (a+b+c)^2 = 0 => a^2 + b^2 + c^2 = -2(ab + bc + ca) 
=> (a^2+b^2+c^2)^2 = 4(ab+bc+ca)^2 = 4.((ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)) 
= 4.((ab)^2+(bc)^2+(ca)^2) 
=> a^4 + b^4 + c^4 = 2.((ab)^2+(bc)^2+(ca)^2) 
Thay lại vào (*) ta có a^4 + b^4 + c^4 = 1/2 

(a+b+c)²=a²+b²+c²+2ac+2bc+2ab

=>0²=1+2(ac+bc+ab)

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

=>(ac+bc+ab)²=a²b²+b²c²+a²c²+2a²bc+2b²ac+2c²ab

(ac+bc+ab)²=a²b²+b²c²+a²c²+2abc(a+b+c)

=>(-1/2)²=a²b²+b²c²+a²c²+2abc.0

=>a²b²+b²c²+a²c²=1/4

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

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

1²=a⁴+b⁴+c⁴+2.1/4

1=a⁴+b⁴+c⁴.1/2

a⁴+b⁴+c⁴=1-1/2=1/2

t.i.c.k mik mik t.i.c.k lại

a) Ta có C = 1 + 4 + 4² + ... + 4⁶

              4C = 4( 1 + 4 + 4² + ... + 4⁶ )

                    = 4 + 4² + 4³ + ... + 4⁷

b) Ta có C = 1 + 4 + 4² + ... + 4⁶

              4C = 4 + 4² + 4³ + ... + 4⁷

⇒ 4C - C = ( 4 + 4² + 4³ + ... + 4⁷ ) - ( 1 + 4 + 4² + ... + 4⁶ )

⇒ 3C = 4 + 4² + 4³ + ... + 4⁷ - 1 - 4 - 4² - ... - 4⁶

⇒ 3C = 4⁷ - 1

⇒ C = \(\dfrac{4^7-1}{3}\) ( đpcm )

vãi tui đang mắc ở bài này 

Bình phương 2 vế a+b+c=0, tính được ab+bc+ca=-1/2.

Bình phương 2 vế ab+bc+ca=-1/2, tính được (ab)²+(bc)²+(ca)²=1/4

Bình phương 2 vế a²+b²+c²=1, ta có:

                  a⁴+b⁴+c⁴+2[(ab)²+(bc)²+(ac)²]=1

           <=> a⁴+b⁴+c⁴+1/2=1

           <=> M=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 ...

\(a,A+B-C=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)

\(=\left(16x^4-15x^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)+\left(-9y^4-6y^4-17y^4\right)-1\)

\(=x^4-10x^3y-x^2y^2-32y^4-1\)

\(b,A-C+B=A+B-C\) ( giống câu a )

\(a,\)

\(A+B+C\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-\left(5x^3y+3x^2y^2+17y^4+1\right)\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)

\(=\left(16x^4-15x^4\right)+\left(-9y^4-6y^4-17y^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)-1\)

\(=x^4-32y^4-10x^3y-x^2y^2-1\)

\(b,\)

\(A-C+B=A+B-C=x^4-32y^4-10x^3y-x^2y^2-1\)

a: A+B-C

=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-C

=x^4-5x^3y+2x^2y^2-15y^4-5x^3y-3x^2y^2-17y^4-1

=x^4-10x^3y-x^2y^2-32y^4-1

b: A-C+B=A+B-C=x^4-10x^3y-x^2y^2-32y^4-1

ta có:

(a+b+c)²=a²+b²+c²+2ab+2bc+2ac

<=>(a+b+c)²=a²+b²+c²+2.(ab+bc+ac)

=>0²     =       1      +2.(ab+bc+ac)

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

(ab+bc+ac)²=a²b²+a²c²+b²c²+ab²c+a²bc+abc²

<=>(ab+bc+ac)²=a²b²+a²c²+b²c²+abc.(a+b+c)

=> (-1/2)²=a²b²+a²c²+b²c²+abc.0

=>a²b²+a²c²+b²c²=1/4

suy ra:

(a²+b²+c²)²=a⁴+b⁴+c⁴+a²b²+a²c²+b²c²

=>1²=a⁴+b⁴+c⁴+1/4

=>a⁴+b⁴+c⁴=1-1/4=3/4

3/4 bạn nhé

3/4 NHA BN

ta có:

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

<=>(a+b+c)^2=a^2+b^2+c^2+2.(ab+bc+ac)

=>0^2      =       1      +2.(ab+bc+ac)

=>ab+bc+ac = -1/2 (ab+bc+ac)2=a2b 2+a2c 2+b2c 2+ab2c+a2bc+abc2

<=>(ab+bc+ac)2=a2b 2+a2c 2+b2c 2+abc.(a+b+c)

=> (-1/2)2=a2b 2+a2c 2+b2c 2+abc.0 =>a2b 2+a2c 2+b2c 2=1/4

suy ra:

(a2+b2+c2 ) 2=a4+b4+c4+a2b 2+a2c 2+b2c 2

=>12=a4+b4+c4+1/4

=>a4+b4+c4=1-1/4=3/4

:A