Question

помогите пожалуйста очень срочно нужно​

Answer 1

№645

1)1249

2)-69370

3)4800

4)4.29

5)0.486875

6)13.3

№646

1)

${x}^{2} - 16 = 0$

${x}^{2} = 16$

$x1 = - 4$

$x2 = 4$

2)

$\frac{1}{9} - {x}^{2} = 0$

$1 - 9 {x}^{2} = 0$

$- 9{x}^{2} = - 1$

${x}^{2} = \frac{1}{9}$

$x1 = - \frac{1}{3}$

$x2 = \frac{1}{3}$

3)

${y}^{2} - 0.25 = 0$

$4 {y}^{2} - 1 = 0$

$4 {y}^{2} = 1$

${y}^{2} = \frac{1}{4}$

$y 1= - 0.5$

$y2 = 0.5$

4)

$4 {x}^{2} - 9 = 0$

$4 {x}^{2} = 9$

${x}^{2} = \frac{9}{4}$

$x1 = - 1.5$

$x2 = 1.5$

№647

1)

${c}^{4} - {m}^{6} = ( {c}^{2} - {m}^{3} ) \times ({c}^{2} + {m}^{3} )$

2)

${p}^{8} - {a}^{10 } = ( {p}^{4} - {a}^{4} ) \times ( {p}^{4} + {a}^{4} )$

3)

${a}^{6} - 9 {m}^{4} = ( {a}^{3} - 3 {m}^{2} ) \times ( {p}^{4} + {a}^{4} )$

4)

${100a}^{6} - 25 {x}^{8} = 25 \times (2 {a}^{3} - {x}^{4}) \times (2 {a}^{3} + {x}^{4})$

5)

$0.49 - {m}^{4} {p}^{12} = ( 0.7 - {m}^{2} {p}^{6} ) \times ( 0.7 + {m}^{2} {p}^{6} )$

6) 36х^2 c ^14-0.16d^4=(6 с^7х-0.4d^2)×(6 с^7х-0.4d^2)

7,8,9 выше