Question

Даю 50 баллов: 1 уровень : 1. Представьте в виде многочлена: а) (а – 5)2; б) (х + 4)2; в) (х – 3)(х + 3). 2. Разложите на множители: а) 4а 2 + 4ав + в2 ; б) а2 – 12а + 36; в) 4х2 – 9 . 3. Вычислите: 152 - 132 2 уровень : 1. Представьте в виде многочлена: а) (2а – 5в)2; б) (3х + у)2; в) (2в + а)(а – 2в). 2. Разложите на множители: а) 4а 2 + 4ав + в2 ; б) а2 – 12а + 36; 3. Вычислите: а) 142 - 132 ; б)152 + 112. в) 4х2 – 9 . 3 уровень : 1. Представьте в виде многочлена: а) (– 5 + х)2; б) (0,1х – 3)(0,1х + 3); в) (0,1у – 0,5)2; г) (– а – 5)2. 2. Разложите на множители: а) – 9х2 + 25; б) 0,04х2 – 9; в) 0,16у2 – 0,4у + 0,25; г) а2 + 2а + 1. 3. Вычислите: 12/13^2-11^2

Answer 1

Ответ:

1-уровень

1.

а) х2 - 10а + 25

б) х2 + 8х + 16

в) х2 - 9

2.

а) (2а+в)2

б) (а-6)2

в) (2х-3)*(2х+3)

3.

225 - 169 = 56

2-уровень

1.

а) 4а2 - 20ав + 25в2

б) 9х2 + 6ху + у2

в) а2 - 4в2

2.

а) (2а+в)2

в) (а-6)2

3.

а) 27

б)346

в) 7

3уровень

1.

а) х2 - 10х + 25

б) 0,01х2 - 9

в) 0,01у2 - 0,1у + 0,25

2.

а) (5-3х)(5+3х)

б) 0,04(х-15)(х+15)

в) 0,01*(4у-5)2

г) (а+1)2

3.

это я не поняла

Объяснение:

надеюсь помогла

Answer 2

1 уровень:

а) (а – 5)² To expand this expression, we can use the formula for squaring a binomial: (a - b)² = a² - 2ab + b². Applying this formula to the expression (а – 5)², we have: (а – 5)² = а² - 2 * а * 5 + 5² = а² - 10а + 25

б) (х + 4)² Applying the same formula, we have: (х + 4)² = х² + 2 * х * 4 + 4² = х² + 8х + 16

в) (х – 3)(х + 3) To expand this expression, we can use the formula for multiplying binomials: (a + b)(a - b) = a² - b². Applying this formula to the expression (х – 3)(х + 3), we have: (х – 3)(х + 3) = х² - 3² = х² - 9

2 уровень:

а) 4а² + 4ав + в² This expression cannot be factored further as it is already in its simplest form.

б) а² – 12а + 36 This expression can be factored as a perfect square trinomial. The formula for factoring a perfect square trinomial is (a - b)² = a² - 2ab + b². Applying this formula to the expression а² – 12а + 36, we can see that it is equal to (а - 6)².

в) 4х² – 9 This expression is a difference of squares and can be factored as (2х + 3)(2х - 3).

3 уровень:

а) (2а – 5в)² Applying the formula for squaring a binomial, we have: (2а – 5в)² = (2а)² - 2 * (2а) * (5в) + (5в)² = 4а² - 20ав + 25в²

б) (3х + у)² Applying the same formula, we have: (3х + у)² = (3х)² + 2 * (3х) * у + у² = 9х² + 6ху + у²

в) (2в + а)(а – 2в) Applying the formula for multiplying binomials, we have: (2в + а)(а – 2в) = а² - (2в)² = а² - 4в²

г) (–а – 5)² Applying the formula for squaring a binomial, we have: (–а – 5)² = (–а)² - 2 * (–а) * 5 + 5² = а² + 10а + 25

2 уровень:

а) –9х² + 25 This expression cannot be factored further as it is already in its simplest form.

б) 0.04х² – 9 This expression can be factored by factoring out a common factor (0.04): 0.04х² – 9 = 0.04(х² – 225) = 0.04(х + 15)(х - 15)

в) 0.16у² – 0.4у + 0.25 This expression cannot be factored further as it is already in its simplest form.

г) а² + 2а + 1 This expression is a perfect square trinomial and can be factored as (а + 1)².

3 уровень:

а) (–5 + х)² Applying the formula for squaring a binomial, we have: (–5 + х)² = (–5)² - 2 * (–5) * х + х² = 25 + 10х + х²

б) (0.1х – 3)(0.1х + 3) Applying the formula for multiplying binomials, we have: (0.1х – 3)(0.1х + 3) = (0.1х)² - (3)² = 0.01х² - 9

в) (0.1у – 0.5)² Applying the formula for squaring a binomial, we have: (0.1у – 0.5)² = (0.1у)² - 2 * (0.1у) * 0.5 + (0.5)² = 0.01у² - 0.01у + 0.25

г) (–а – 5)² This expression was already evaluated in the previous level and is equal to а² + 10а + 25.

Вычислите:

To compute the expressions, we need the values of the variables provided. Could you please provide the values for а, в, х, у in the expressions you mentioned?