Question

(4×|x-1|+1/2)^2=11×(x-1)^2+5/4

Answer 1

Ответ: 1 1/5,  4/5

Пошаговое объяснение:

16*|x-1|^2+2*4*|x-1|*1/2+1/4=11*|x-1|^2+5/4,  5*|x-1|^2+4*|x-1|-1=0, пусть |x-1|=y, y>=0,  тогда 5y^2+4y-1=0,  D=36,  y= -4-6 /10=-1  не удовл.,  у=1/5,

|x-1|=1/5,  x-1=1/5,  x=1 1/5,  x-1=-1/5,  x=1-1/5=4/5,  использовали ф-лу

(а+в)^2=a^2+2ab+b^2),  (x-1)^2=|x-1|^2

Answer 2

To solve the equation (4×|x-1|+1/2)^2 = 11×(x-1)^2 + 5/4, we need to follow these steps:

Step 1: Simplify both sides of the equation. Step 2: Solve for x.

Let's start with Step 1:

(4×|x-1| + 1/2)^2 = 11×(x-1)^2 + 5/4

Step 1: Simplify both sides: Expanding the square on the left side: (4×|x-1|)^2 + 2×(4×|x-1|)×(1/2) + (1/2)^2 = 11×(x-1)^2 + 5/4

Simplify: 16×|x-1|^2 + 2×(2×|x-1|) + 1/4 = 11×(x-1)^2 + 5/4

Now, simplify the absolute value expression: Remember that |a| = a if a ≥ 0 and |a| = -a if a < 0.

Case 1: (x - 1) ≥ 0 |x - 1| = x - 1

Case 2: (x - 1) < 0 |x - 1| = -(x - 1) = -x + 1

So, we have two cases to consider:

Case 1: (x - 1) ≥ 0 The equation becomes: 16×(x - 1)^2 + 4×(x - 1) + 1/4 = 11×(x - 1)^2 + 5/4

Case 2: (x - 1) < 0 The equation becomes: 16×(-x + 1)^2 + 4×(-x + 1) + 1/4 = 11×(-x + 1)^2 + 5/4

Step 2: Solve for x in both cases.

Case 1: (x - 1) ≥ 0 16×(x - 1)^2 + 4×(x - 1) + 1/4 = 11×(x - 1)^2 + 5/4

Let's solve this equation:

16×(x - 1)^2 + 4×(x - 1) + 1/4 = 11×(x - 1)^2 + 5/4

Expand and simplify: 16×(x^2 - 2x + 1) + 4x - 4 + 1/4 = 11×(x^2 - 2x + 1) + 5/4

16x^2 - 32x + 16 + 4x - 4 + 1/4 = 11x^2 - 22x + 11 + 5/4

Combine like terms: 16x^2 - 32x + 4x + 16 - 4 + 1/4 = 11x^2 - 22x + 11 + 5/4

16x^2 - 28x + 13/4 = 11x^2 - 22x + 16/4 + 5/4

16x^2 - 28x + 13/4 = 11x^2 - 22x + 21/4

Move all terms to one side of the equation: 16x^2 - 11x^2 - 28x + 22x + 13/4 - 21/4 = 0

5x^2 - 6x - 8/4 = 0

5x^2 - 6x - 2 = 0

Now, you can use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 5, b = -6, and c = -2

x = (6 ± √((-6)^2 - 4×5×(-2))) / 2×5

x = (6 ± √(36 + 40)) / 10

x = (6 ± √76) / 10

x = (6 ± √(4×19)) / 10

x = (6 ± 2√19) / 10

x = (3 ± √19) / 5

So, in Case 1, the solutions for x are: x = (3 + √19) / 5 and x = (3 - √19) / 5

Now, let's proceed to Case 2: (x - 1) < 0

16×(-x + 1)^2 + 4×(-x + 1) + 1/4 = 11×(-x + 1)^2 + 5/4

Expand and simplify: 16×(x^2 - 2x + 1) - 4x + 4 + 1/4 = 11×(x^2 - 2x + 1) + 5/4

16x^2 - 32x + 16 - 4x + 4 + 1/4 = 11x^2 - 22x + 11 + 5/4

16x^2 - 36x + 17/4 = 11x^2 - 22x + 16/4 + 5/4

16x^2 - 36x + 17/4 = 11x^2 - 22x + 21/4

Move all terms to one side of the equation: 16x^2 - 11x^2 - 36x + 22x + 17/4 - 21/4 = 0

5x^2 - 14x - 4/4 = 0

5x^2 - 14x - 1 = 0

Using the quadratic formula again:

x = (14 ± √((-14)^2 - 4×5×(-1))) / 2×5

x = (14 ± √(196 + 20)) / 10

x = (14 ± √216) / 10

x = (14 ± √(36×6)) / 10

x = (14 ± 6√6) / 10

x = (7 ± 3√6) / 5

So, in Case 2, the solutions for x are: x = (7 + 3√6) / 5 and x = (7 - 3√6) / 5

Therefore, the complete solution to the equation is: x = (3 + √19) / 5, x = (3 - √19) / 5, x = (7 + 3√6) / 5, and x = (7 - 3√6) / 5.