Question

помогите с алгеброй (хотя бы 3 задание)​

Answer 1

$1.$

$1) \: \: \: \: \: { {(2x}^{2}) }^{3} \times ( - 0.75x {y}^{2} )^{3} = {8x}^{6} \times ( - 0,421875 {x}^{3} {y}^{6} ) = 3.375 {x}^{9} {y}^{6}$

$2) \: \: \: \: \: 2x - 3y - (x + y) = 2x - 3y - x - y = x - 4y$

$2.$

$1) \: \: \: \: \: 0.01 {a}^{4} \times {b}^{2} = 0.01 {a}^{4} {b}^{2} \\ 2) \: \: \: \: \: 2 \frac{1}{4} {x}^{6} \times {y}^{4} = 2.5 {x}^{6} {y}^{4} \: \: \:$

$3.$

$1) \: \: \: \: \: 7x(4x - y) + 4x(x - 7y) = 28 {x}^{2} - 7xy + {4x}^{2} - 28xy = {32x}^{2} - 35xy$

$2) \: \: \: \: \: 2ab - (a + b)(3a - b) = 2ab - {3a}^{2} + ab - 3ab + {b}^{2} = {b}^{2} - {3a}^{2}$

$4.$

$1) \: \: \: \: \: 7 {x}^{2} {y}^{2} - 14x {y}^{3} + 28 {x}^{2} {y}^{4} = 7x {y}^{2}(x - 2 {y}+ 4x {y}^{2} )$

$2) \: \: \: \: \: 24 {p}^{2} {q}^{2} - 16p {q}^{3} = 8p {q}^{2} (3p - 2q)$

$5.$

$(x - 1)(x + 3) = x(x - 2) \\ {x}^{2} + 3x - x - 3 = {x}^{2} - 2x \\ 3x - x + 2x = 3 \\ 4x = 3 \\ x = \frac{3}{4}$