Question

Help помогите
$8 \frac{2}{3} {x}^{2} - 9ay {}^{2} - 9x {}^{3} \\ \\ \\ 27 {a}^{2}z - 24.89 {a}^{2} z + 3 \frac{1}{5} {y}^{2} - 15 {y}^{2} \\ \\ \\ 3.12ab + 7 \frac{5}{6} {m}^{3} - 4 \frac{1}{6} {m}^{3} + 16.82ab$
​

Answer 1

1)

$8 \frac{2}{3} {x}^{3} - 16a {y}^{2} + 9a {y}^{2} - 9 {x}^{3} = 8\frac{2}{3} {x}^{3} - 9 {x}^{3} - 16a {y}^{2} + 9a {y}^{2} = ( 8\frac{2}{3} - 9) {x}^{3} - (16 - 9)a {y}^{2} = ( 8\frac{2}{3} - 8 \frac{3}{3} ) {x}^{3} - 7a {y}^{2} = - \frac{1}{3} {x}^{3} - 7a {y}^{2}$

2)

$27 {a}^{2} z - 24.89 {a}^{2} z + 3 \frac{1}{5} {y}^{2} - 15 {y}^{2} = (27 - 24.89) {a}^{2} z - (15 - 3 \frac{1}{5} ) {y}^{2} = 2.11 {a}^{2} z - (15 - 3 \frac{2}{10} ) {y}^{2} = 2.11 {a}^{2} z - (15 - 3.2) {y}^{2} = 2.11 {z}^{2} - 11.8 {y}^{2}$

3)

$3.12ab + 7 \frac{5}{6} {m}^{3} - 4 \frac{1}{6} {m}^{3} + 16.82ab = (3.12 + 16.82)ab + (7 \frac{5}{6} - 4 \frac{1}{6} ) {m}^{3} = 19.94ab + ((7 + \frac{5}{6}) - (4 + \frac{1}{6} )) {m}^{3} = 19.94ab + ((7 - 4) + ( \frac{5}{6} - \frac{1}{6} )) {m}^{3} = 19.94ab + (3 + \frac{4}{6} ) {m}^{3} = 19.94ab + (3 + \frac{2}{3} ) {m}^{3} = 19.94ab + 3 \frac{2}{3} {m}^{3}$

4)

$19.8 {x}^{2} - 30 \frac{1}{9} kt + 31kt - 20 {x}^{2} = - (20 - 19.8) {x}^{2} + (31 - 30 \frac{1}{9} )kt = - 0.2 {x}^{2} + (30 \frac{9}{9} - 30 \frac{1}{9} )kt = - 0.2 {x}^{2} + ((30 - 30) + ( \frac{9}{9} - \frac{1}{9} ))kt = - 0.2 {x}^{2} + \frac{8}{9} kt$