Докажите тождество

(7 х - 3) (4-8 х) + 2 х (28 х - 26) = - 12

(1.1X^2) (X^2) - 10) - x (1.1X^3-9X) = - 2X^2

(-y^3 + 5y) 2y - 10y^2 (1+0.2y^2) = - 4y^4

(2.5a + b^2) (-4a) + 2a (5a-b^2) = - 6ab^2

1) (7x-3) (4-8x) + 2x (28x-26) = - 12

28x-12-56x^2+24x + 56x^2-52x=-12

-12=-12

2) 1,1x^2 (x^2-10) - X (1,1X^3-9x) = - 2x^2

1,1x^4-11x^2-1,1x^4+9x^2=-2x^2

-2x^2 = - 2x^2

3) (-y^3+5y) 2y-10y^2 (1+0,2y^2) = - 4y^4

-2y^4+10y^2-10y^2-2y^4=-4y^4

-4y^4=-4y^4

4) (2.5a + b^2) (-4a) + 2a (5a-b^2) = - 6ab^2

-10a^2-4ab^2+10a^2-2ab^2=-6ab^2

-6ab^2=-6ab^2

1.28 х - 56 х²-12+24x+56x²-52x=-12

2. непонятно

3. - 2y^4+10y²-10y²-2y^4=-2y^4-2y^4=4y^4

4. - 10a²-4ab²+10a²-2ab²=-4ab²-2ab²=-6ab²