Có: \(a^2+b^2=1-2ab\)

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

Mà: \(a>0;b>0\Rightarrow a+b>0\)

Do đó: \(a+b=1\)

Có: \(M=a^3+b^3+3ab=a^3+b^3+3ab\left(a+b\right)=\left(a+b\right)^3=1^3=1\)

Ta có : M=a³+b³+3ab

=(a+b)(a²-ab+b²)+3ab=(a+b)(a²+b²-ab)+3ab

Ma : a²+b²=1-2ab 

\(\Rightarrow\)(a+b)(a²+b²-ab)+3ab

=(a+b)(1-2ab-ab)+3ab

=(a+b)(1-3ab)+3ab

=a+b

​Ma : a và b là hai số dương \(\Rightarrow\)a>0 va b>0
\(\Rightarrow\)Gia tri cua bieu thuc M=a³+b³+3ab = a+b .

\(.\)M= bn ghi lại đề nha ^.^

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=1^3-3ab.1+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2.1\)

\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)

\(M=1-3ab+3ab-6a^2b^2+6a^2b^2\)\(=1\)

k cho mình nha bn thanks nhìu <3 <3       (^3^)

2. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(1)

Đặt \(x^2+5x+4=t\)

(1) = \(t.\left(t+2\right)-24\)

\(=t^2+2t+1-25\)

\(=\left(t+1\right)^2-25\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)(2)

Thay \(t=x^2+5x+4\)vào (2) ta có:

(2) = \(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

k mình nha bn <3 thanks

Easy thật : 

\(a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(=a^3+b^3-\left(a-b\right)^2\left(a-b\right)\)

\(=a^3+b^3-\left(a-b\right)^3\)

Thay \(a=-4;b=4\)vào biểu thức , ta được : 

\(\left(-4\right)^3+4^3-\left(-4-4\right)^3\)

\(=8^3\)

\(=512\)

a³ + b³ - ( a² - 2ab + b² )( a - b )

= a³ + b³ - ( a - b )²( a - b )

= a³ + b³ - ( a - b )³

Thế a = -4 ; b = 4 ta được

(-4)³ + 4³ - ( -4 - 4 )³

= -64 + 64 - ( -512 )

= 512

a, chắc bạn chép nhầm đề rồi đó nếu mà là 3ab thì k làm đc đâu

M=a³ + a² - b3 + b2 + 3ab2 -2ab +3ab2

= (a-b)3 +(a-b)2

= 343+49=392

b, P= -(3x+4x2+1/4x-2014)

= - [ (2x)2 -4x+1 +x +1/4x - 2015]

= -[ (2x-1)2- (2x-1)2/4x +1 -2015]

Max P = 2014   X=1/2

Ta có: \(a+b+c=0\)

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

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

\(\Leftrightarrow1+2\left(ab+bc+ca\right)=0\)

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

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

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

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

Thay vào ta được:

\(A=a^4+b^4+c^4\)

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

\(A=1-\frac{1}{2}=\frac{1}{2}\)

Từ \(a+b+c=0\)

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

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

Vì \(a^2+b^2+c^2=1\)

\(\Rightarrow1+2\left(ab+bc+ca\right)=0\)

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

\(\Leftrightarrow ab+bc+ca=\frac{-1}{2}\)

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

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+b^2ac+c^2ab\right)=\frac{1}{4}\)

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

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

Ta có: \(a^2+b^2+c^2=1\)

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

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

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

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

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

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

hay \(A=a^4+b^4+c^4=\frac{1}{2}\)

ý a)

(a+b)^2=a^2+b^2+2ab

=> 529=a^2+b^2+246  => a^2+b^2=283

(a^2+b^2)^2=a^4+b^4+2.a^2.b^2

=> 80089=a^4+b^4+30258   => a^4+b^4=49831

(a^2+b^2)(a^4+b^4)=a^6+b^6+a^2.b^4+b^2.a^4=a^6+b^6+a^2.b^2.(a^2+b^2)

=> 14102173=a^6+b^6+15129.283  => a^6+b^6=9820666

còn lại bạn tự tính

ý b)

(x+y)^3=x^3+y^3+3xy.(x+y)

suy ra x^3+y^3+3xy=1

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

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

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

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

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

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

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