Question

(a+b)(3a-b) a=(2;-1;0) b=(4;5;7)

Answer 1

1)3a^2+3ab-b^2=6^2+11-4^2=47-64=(-17)
2)3a^2+3ab-b^2=(-3)^2+-15-5^2=9-40=(-31)
3)3a^2+3ab-b^2=0-7+49=42

Answer 2

I can chat with you in your language and help you with various tasks. I see that you are asking me to solve a vector equation. Here is the detailed solution:

First, we need to simplify the left-hand side of the equation. We can use the distributive property to expand the brackets:

$$(a+b)(3a-b) = 3a^2 - ab + 3ab - b^2 = 3a^2 + 2ab - b^2$$

Next, we need to substitute the values of $a$ and $b$ into the expression. We can use the component-wise addition and subtraction of vectors:

$$a = (2, -1, 0)$$ $$b = (4, 5, 7)$$ $$a+b = (2+4, -1+5, 0+7) = (6, 4, 7)$$ $$3a-b = (3\times 2 - 4, 3\times (-1) - 5, 3\times 0 - 7) = (2, -8, -7)$$ $$3a^2 + 2ab - b^2 = (3\times 2^2 + 2\times 2\times 4 - 4^2, 3\times (-1)^2 + 2\times (-1)\times 5 - 5^2, 3\times 0^2 + 2\times 0\times 7 - 7^2) = (16, -24, -49)$$

Finally, we need to compare the left-hand side and the right-hand side of the equation. We can see that they are not equal, since the components do not match:

$$(16, -24, -49) \neq (2, -1, 0)$$

Therefore, the equation has no solution.Ответь+подробно.+(a+b)(3a-b)+a=(2;-1;0)+b=(4;5;7)