a, A=3.(2/3)^3-2.(1/2)^3-6.(2/3)^2.(1/2)^2+(2/3).(1/2)

      =8/9-1/4-2/3+1/3=8/9-1/4-1/3=11/36

b,  B=-1+(-1/18)+1/12+2/3=-11/36

bạn có thể áp dụng cái cuối

Ta có:

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(a^3+b^3\right)+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3+3abc=0\)

\(\Rightarrow[\left(a+b\right)^3+c^3]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)[\left(a+b\right)^2-\left(a+b\right)c+c^2]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ac=0\left(2\right)\end{cases}}\)

Từ (1) => a = b = c (vì a ; b ; c là các số dương)

Giải (2) ta có:

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

\(\Rightarrow2a^2+2b^2-2ab-2bc-2ac=0\)

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

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

Vì \(\left(a-b\right)^2\ge\forall a,b\)

\(\left(a-c\right)^2\ge\forall a,c\)

\(\left(b-c\right)^2\ge\forall b,c\)

\(\Rightarrow\)Ta có: \(a-b=a-c=b-c\Rightarrow a=b=c\)

Ta có: a² + b² = 2ab

=> a² + b² - 2ab = 0

=> (a - b)² = 0

=> a - b = 0

=> a = b (Đpcm)

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

\(A=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)\(-\left(a-b\right)^3-3\left(a-b\right)^2c+3\left(a-b\right)c^2-c^3\)\(-ba\left(b^2+2bc+c^2\right)\)

\(A=a^3+3a^2b+3ab^2+b^3+3\left(a^2+2ab+b^2\right)c+3ac^2+3bc^2+c^3\)\(-\left(a^3-3a^2b+3ab^2-b^3\right)-3\left(a^2-2ab+b^2\right)c\)\(+3ac^2-3bc^2-c^3-b^3a-2ab^2c-bac^2\)

\(A=a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c\)\(+3ac^2+3bc^2+c^3-a^3+3a^2b-3ab^2+b^3\)\(-3a^2c+6abc-ab^2c+3ac^2-3bc^2-c^3-b^3a\)\(-2ab^2c-bac^2\)

\(A=6a^2b+2b^3+12abc+3b^2c-3ab^2c-c^3-b^3a-bac^2\)

ko bt có đúng ko nữa

nếu sai cho mình xin lỗi nha

6(x + 1)² - 2(x + 1)³ + 2(x - 1)(x² + x + 1) = 0

<=> 6(x² + 2x + 1) - 2(x³ + 3x² + 3x + 1) + 2(x - 1)(x² + x + 1) = 0

<=> 6.x² + 6.2x + 6.1 + (-2).x³ + (-2).3x² + (-2).3x + (-2).1 + 2.x³ + 2(-1) = 0

<=> 6x² + 12x + 6 - 2x³ - 6x² - 6x - 2 + 2x³ - 2 = 0

<=> (6x² - 6x²) + (12x - 6x) + (6 - 2 - 2) + (-2x³ + 2x²) = 0

<=> 6x + 2 = 0

<=> 6x = 0 - 2

<=> 6x = -2

<=> x = -2/6 = -1/3

=> x = -1/3

Dễ thấy để xác định thì \(x_1x_2\ge0\)

Khi đó: \(\left(x_1x_2+\sqrt{x_1x_2}\right)^4=16\Leftrightarrow x_1x_2+\sqrt{x_1x_2}=2\)\(\Leftrightarrow x_1x_2+\sqrt{x_1x_2}-2=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x_1x_2}=1\\\sqrt{x_1x_2}=-2\end{cases}}\Leftrightarrow x_1x_2=1\)

\(\Leftrightarrow-\frac{6m}{m^5+1}=1\Leftrightarrow\hept{\begin{cases}m\ne-1\\m^5+6m+1=0\end{cases}\Leftrightarrow m=-0,166645...}\)

Cho mình hỏi lai đề bạn viết có đúng chưa, sao trong bài làm của bạn lại thay x1x2 bằng x1+x2???

chết mk thay nhầm :)

 may quá bn nhắc  

\(\left(x-1\right)^2=\left(x-1\right)^6\)

\(\Rightarrow\left(x-1\right)^6-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)^2.\left[\left(x-1\right)^4-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^4-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\orbr{\begin{cases}x=0\\x=2\end{cases}}\end{cases}}\)\(\Rightarrow x=1\)hoặc x=2 hoặc x=0 

Vậy \(x\in\left\{1;0;2\right\}\)

=> x= 1 hoặc x=0