Question

1)7-5х+х²=02)169-х²=03)-0,1х²+10=04)7х+13-6х²=0​

Answer 1

Ответ:

1)x≠R

2)169-x²=0

-x²=-169

x²=169

x1=-13

x2=13

3)-0,1x²+10=0

-х²+100=0

-х²=-100

х²=100

x1=-10

x2=10

4)7x+13-6x²=0

-7x-13x+6x²=0

6x²-7x-13=0

6x²+6x-13x-13=0

6x×(x+1)-13(x+1)=0

(х+1)(6х-13)=0

х+1=0

6х-13=0

х1=-1

х2=13/6

Answer 2

It looks like you've provided four quadratic equations. If you'd like, I can help you solve them. Here are the solutions for each equation:

1. $7 - 5x + x^2 = 0$: This is a quadratic equation in standard form. You can solve it by factoring or using the quadratic formula. Factoring: $(x - 7)(x + 1) = 0$ Solutions: $x = 7$ or $x = -1$

2. $169 - x^2 = 0$: This is also a quadratic equation, but it's already in a special form known as the difference of squares. Factoring: $(13 - x)(13 + x) = 0$ Solutions: $x = 13$ or $x = -13$

3. $-0.1x^2 + 10 = 0$: This quadratic equation involves decimals. To solve it, you can multiply the whole equation by 10 to eliminate the decimal. Multiply by 10: $-x^2 + 100 = 0$ Subtract 100: $-x^2 = -100$ Divide by -1: $x^2 = 100$ Take square roots: $x = \pm 10$ Solutions: $x = 10$ or $x = -10$

4. $7x + 13 - 6x^2 = 0$: This is another quadratic equation. You can rearrange it to solve for $x$. Combine like terms: $-6x^2 + 7x + 13 = 0$ You can solve this using the quadratic formula or factoring. Factoring is a bit more complex in this case, so I'll use the quadratic formula. Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For this equation, $a = -6$, $b = 7$, and $c = 13$. $x = \frac{-7 \pm \sqrt{7^2 - 4(-6)(13)}}{2(-6)}$ $x = \frac{-7 \pm \sqrt{49 + 312}}{-12}$ $x = \frac{-7 \pm \sqrt{361}}{-12}$ $x = \frac{-7 \pm 19}{-12}$ Solutions: $x = \frac{-26}{-12} = \frac{13}{6}$ or $x = \frac{12}{6} = 2$

These are the solutions to the given quadratic equations. If you have any questions or need further assistance, feel free to ask!