Question

√x-1=3x-7 √x^2-7x+21=3

Answer 1

1) ОДЗ :

$\left \{ {{x-1\geq 0} \atop {3x-7\geq 0}} \right.\\\\\left \{ {{x\geq1 } \atop {x\geq2\frac{1}{3}}} \right.$

x ∈ [2 1/3 ; + ∞)

$1)(\sqrt{x-1})^{2}=(3x-7)^{2}\\\\x-1=9x^{2}-42x+49\\\\9x^{2}-42x-x+49+1=0\\\\9x^{2} -43x+50=0\\\\D=(-43)^{2}-4*9*50=1849-1800=49=7^{2}\\\\x_{1}=\frac{43+7}{18}=2\frac{7}{9}\\\\x_{2} =\frac{43-7}{18} =2$

Ответ : 2 7/9

2) ОДЗ :

x² - 7x + 21 ≥ 0

x² - 7x + 21 = 0

D = (- 7)² - 4 * 21 = 49 - 84 = - 35 < 0

Дискриминант меньше нуля, а старший коэффициент равен 1 > 0 , значит  x² - 7x + 21 > 0 при всех действительных значениях x .

$(\sqrt{x^{2}-7x+21 })^{2}=3^{2}\\\\x^{2}-7x+21=9\\\\x^{2}-7x+12=0\\\\x_{1} =4\\\\x_{2}=3$

Ответ : 3 ; 4

Answer 2

$\sqrt{x - 1} = 3x - 7 \\ {( \sqrt{x - 1)} }^{2} = {(3x - 7)}^{2} \\ x - 1 = 9 {x}^{2} - 42x + 49 \\ 9 {x}^{2} - 42 x + 49 - x + 1 = 0 \\ 9 {x}^{2} - 43x + 50 = 0 \\ D = 49 \\ \sqrt{D} = 7 \\ x1 = \frac{43 + 7}{2 \times 9} = \frac{50}{18} = \frac{25}{9} \\ x2 = \frac{43 - 7}{2 \times 9} = 2$
ОДЗ x-1≥0
x≥1
Ответ: 25/9, т.к. x≠2

$\sqrt{ {x}^{2} - 7x + 21} = 3 \\ ({ \sqrt{ {x}^{2} - 7x + 21}) }^{2} = {3}^{2} \\ {x}^{2} - 7x + 21 = 9 \\ {x - 7x}^{2} + 12 = 0 \\D = 1 \\ \sqrt{D} = 1 \\ x1 = \frac{7 + 1}{2} = 4 \\ x2 = \frac{7 - 1}{2} = 3$
ОДЗ x∈R
Ответ: x=3, x=4.

Answer 3

To solve the equations, let's take one step at a time.

Equation 1: √x - 1 = 3x - 7

Step 1: Isolate the square root term by moving the constant term to the other side: √x = 3x - 7 + 1 √x = 3x - 6

Step 2: Square both sides to eliminate the square root: (√x)^2 = (3x - 6)^2 x = (3x - 6)^2

Step 3: Expand and simplify the right side: x = (3x - 6)(3x - 6) x = 9x^2 - 36x + 36

Step 4: Move all terms to one side to set the equation to zero: 9x^2 - 36x + 36 - x = 0 9x^2 - 37x + 36 = 0

Now, we have a quadratic equation. We can use the quadratic formula to find the values of x:

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

where a = 9, b = -37, and c = 36.

Step 5: Plug the values into the quadratic formula and solve for x: x = [37 ± √((-37)^2 - 4 * 9 * 36)] / 2 * 9 x = [37 ± √(1369 - 1296)] / 18 x = [37 ± √73] / 18

So, the two solutions for Equation 1 are: x = (37 + √73) / 18 x = (37 - √73) / 18

Moving on to Equation 2: √(x^2 - 7x + 21) = 3

Step 1: Square both sides to eliminate the square root: (√(x^2 - 7x + 21))^2 = 3^2 x^2 - 7x + 21 = 9

Step 2: Move the constant term to the other side: x^2 - 7x + 21 - 9 = 0 x^2 - 7x + 12 = 0

This is another quadratic equation. We'll use the quadratic formula again:

a = 1, b = -7, and c = 12.

Step 3: Plug the values into the quadratic formula and solve for x: x = [7 ± √((-7)^2 - 4 * 1 * 12)] / 2 * 1 x = [7 ± √(49 - 48)] / 2 x = [7 ± √1] / 2

So, the two solutions for Equation 2 are: x = (7 + 1) / 2 x = (7 - 1) / 2

Now, let's simplify these solutions:

For Equation 1: x = (37 + √73) / 18 ≈ 2.569 x = (37 - √73) / 18 ≈ 1.431

For Equation 2: x = (7 + 1) / 2 = 4 x = (7 - 1) / 2 = 3

Thus, the solutions for the system of equations are: x ≈ 2.569, x ≈ 1.431, x = 4, and x = 3.