Question

Прошу выполнить первые 3 номера,срочно,даю 35 балов.

Answer 1

а) (а + b - 2ab/a+b) × a^2-b^2/a^2+b^2 = (a(a+b)+b(a+b)-2ab/a+b) × a^2-b^2/a^2+b^2 = (a^2+ab+ab+b^2-2ab/a+b) × a^2-b^2/a^2+b^2 = a^2+b^2/a+b × (a-b)(a+b)/a^2+b^2 = a - b

б) (m/mn-n^2 - 1/m-n) ÷ n/n-m = m-n/n(m-n) × n-m/n = n-m/n^2

в) 4x^2-9/9x^2-6x+1 ÷ 2x-3/3x-1 + 4-x/1-3x = (2x-3)(2x+3)/(3x-1)^2 × 3x-1/2x-3 + 4-x/1-3x = 2x-3/3x-1 + 4-x/1-3x = -2x+3+4-x/1-3x = -3x+7/1-3x

Answer 2

Ответ:

a ) ( a + b - 2ab / a + b ) x a ^ 2 - b ^ 2 / a ^ 2 + b ^ 2 ( a ( a + b ) + b ( a + b ) -2ab / a + b ) x a ^ 2 - b ^ 2 / a ^ 2 + b ^ 2 = ( a ^ 2 + ab + ab + b ^ 2-2ab / a + b ) x a ^ 2 - b ^ 2 / a ^ 2 + b ^ 2 = a ^ 2 + b ^ 2 / a + b * ( a - b ) ( a + b ) / a ^ 2 + b ^ 2 = a - b 6 ) ( m / mn - n ^ 2 - 1 / m - n ) = n / n - m = m - n / n ( m - n ) * n - m / n = n - m / n ^ 2 B ) 4x ^ 2-9 / 9x ^ 2-6x + 1 = 2x - 3 / 3x - 1 + 4 - x / 1-3x = ( 2x - 3 ) ( 2x + 3 ) ( 3x - 1 ) ^ 2 * 3x - 1 / 2x - 3 + 4 - x / 1-3x = 2x - 3 / 3x - 1 + 4 - x / 1-3x = -2x + 3 + 4 - x / 1-3x = -3x + 7 / 1-3x