Question

1 задание Представьте выражение в виде квадрата двучлена, если это возможно1) 49x2 +28xy +4y2= 2) 0,01x4 +0,2x2y3 +y6 =3) 121a4 +11a2b2 + b4 =4) 4a2 + 4a +b2=

Answer 1

Ответ:

1)(7x+2y)²

2)(0.1x²+y³)²

4)(2a+b)²

3) если бы у нас был 121а⁴+22а²b²+b⁴, то у нас бы вышел (11а²+b²)², но вместо 22 у нас 11

Answer 2

1. To express the expression 49x^2 + 28xy + 4y^2 as a perfect square trinomial, we can factor it as follows:

49x^2 + 28xy + 4y^2 = (7x + 2y)^2

Therefore, the expression can be written as the square of the binomial (7x + 2y).

2. The expression 0.01x^4 + 0.2x^2y^3 + y^6 cannot be written as a perfect square trinomial since the exponents of x and y are different in each term.

3. Similarly, the expression 121a^4 + 11a^2b^2 + b^4 cannot be expressed as a perfect square trinomial since the exponents of a and b are different in each term.

4. The expression 4a^2 + 4a + b^2 cannot be written as a perfect square trinomial since the coefficients of a^2 and b^2 terms are not equal.