• Предмет: Математика
  • Автор: theskile02
  • Вопрос задан 3 месяца назад

2.10. 1) 1/5 x²- -5=0; 3) 1,44-x²-3x²; 5) (2y-1)²-10-4y; 2) 9y²-6,25-0; 4) ¾ x² = −3,5+x³²; 6) (3m-2)(3m+2)=5m².
help plsss, i cnt understnd ths

Приложения:

Ответы

Ответ дал: alesyaperova
0

1) \frac{1}{5}  {x}^{2}  = 0 + 5 \\  \frac{1}{5}  {x}^{2}  = 5 \\  {x}^{2} = 5 \div  \frac{1}{5}  = 5 \times  \frac{5}{1}  \\  {x}^{2}  = 25 \\ x1 = 5 \\ x2 =  - 5

2)9 {y}^{2}  - 6.25 = 0 \\ 9 {y}^{2}  = 0 + 6.25 \\ 9 {y}^{2}  = 6.25 \\  {y}^{2}  = 6.25 \div 9 \\  {y}^{2}   = 6 \frac{25}{100}  \div 9 = 6 \frac{1}{4}  \div 9 \\  {y}^{2}  =  \frac{25}{4}  \times  \frac{1}{9}  \\  {y}^{2}  =  \frac{25}{36}

y1 =  \frac{5}{6}  \\ y2 =  -  \frac{5}{6}

3)1.44 -  {x}^{2}  = 3 {x}^{2}  \\ 1.44 - 4 {x}^{2}  = 0 \\ 4 {x}^{2}  = 1.44 + 0 \\ 4 {x}^{2}  = 1.44 \\  {x}^{2}  = 1.44 \div 4 \\  {x}^{2}  = 0.36 \\ x1 = 0.6 \\ x2 =  - 0.6

4) \frac{5}{7}  {x}^{2}  =  - 3.5 +  {x}^{2}  \\  \frac{5}{7}  {x}^{2}  -  {x}^{2}  + 3.5 = 0 \\  -  \frac{2}{7}  {x}^{2}  + 3.5 = 0 \\  -  \frac{2}{7} {x}^{2}   = 0 - 3.5 \\  -  \frac{2}{7}  {x}^{2}  =  - 3.5 \\  {x }^{2}  =  - 3.5 \div ( -  \frac{2}{7} ) \\

 {x}^{2}  =  - 3 \frac{5}{10}  \times ( -  \frac{7}{2} ) \\  {x}^{2}  =  - 3 \frac{1}{2}  \times( -   \frac { 7}{2})  \\  {x}^{2}  =   - \frac{7}{2}  \times ( -  \frac{7}{2} ) \\  {x }^{2}  =  \frac{49}{4}  \\ x1 =  \frac{7}{2}  = 3.5 \\ x2 =  \frac{7}{2}  =  - 3.5

5)(2y - 1) {}^{2}  = 10 - 4y \\  {4y}^{2}  - 4y + 1 = 10 - 4y \\ 4 {y}^{2}  - 4y + 1 - 10 + 4y = 0 \\ 4 {y}^{2}  - 9 = 0 \\  {4y}^{2}  = 0 + 9 \\  {4y}^{2}  = 9 \\  {y}^{2}  = 9 \div 4 \\  {y}^{2}  =  \frac{9}{4}  \\ y1 =  \frac{3}{2}  = 1.5 \\

y2 =  - 1.5

6)(3m - 2)(3m + 2) = 5 {m}^{2}  \\  {9m}^{2}  - 4 - 5 {m}^{2}  = 0 \\  {4m}^{2} - 4 = 0 \\  {4m}^{2}  = 0 + 4 \\  {4m}^{2}  = 4 \\  {m}^{2}  = 4 \div 4 \\  {m}^{2}  = 1 \\ m1 = 1 \\ m2 =  - 1

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