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Sagot :
Sure, let's solve each equation step by step.
### Equation 1:
[tex]\[ \frac{2 x}{x+1} + \frac{5}{2 x} = 2 \][/tex]
First, let's find a common denominator, which is [tex]\(2x(x+1)\)[/tex]:
[tex]\[ \frac{(2x)^2 + 5(x + 1)}{2x(x + 1)} = 2 \][/tex]
Simplify the numerator:
[tex]\[ \frac{4x^2 + 5x + 5}{2x(x + 1)} = 2 \][/tex]
Cross-multiply to clear the fraction:
[tex]\[ 4x^2 + 5x + 5 = 4x(x + 1) \][/tex]
Expand the right-hand side:
[tex]\[ 4x^2 + 5x + 5 = 4x^2 + 4x \][/tex]
Subtract [tex]\(4x^2\)[/tex] from both sides:
[tex]\[ 5x + 5 = 4x \][/tex]
Subtract [tex]\(4x\)[/tex] from both sides:
[tex]\[ x + 5 = 0 \][/tex]
Therefore, solving for [tex]\(x\)[/tex]:
[tex]\[ x = -5 \][/tex]
So the solution to the first equation is:
[tex]\[ x = -5 \][/tex]
### Equation 2:
[tex]\[ \frac{5}{v} + \frac{5}{v + 10} = \frac{4}{3} \][/tex]
Find a common denominator, which is [tex]\(v(v + 10)\)[/tex]:
[tex]\[ \frac{5(v + 10) + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]
Simplify the numerator:
[tex]\[ \frac{5v + 50 + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]
Combine like terms:
[tex]\[ \frac{10v + 50}{v(v + 10)} = \frac{4}{3} \][/tex]
Cross-multiply to clear the fraction:
[tex]\[ 3(10v + 50) = 4v(v + 10) \][/tex]
Expand both sides:
[tex]\[ 30v + 150 = 4v^2 + 40v \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 4v^2 + 40v - 30v - 150 = 0 \][/tex]
[tex]\[ 4v^2 + 10v - 150 = 0 \][/tex]
Divide the entire equation by 2 to simplify:
[tex]\[ 2v^2 + 5v - 75 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (2v + 15)(v - 5) = 0 \][/tex]
So, the solutions for [tex]\(v\)[/tex] are:
[tex]\[ 2v + 15 = 0 \implies v = -\frac{15}{2} = -7.5 \][/tex]
[tex]\[ v - 5 = 0 \implies v = 5 \][/tex]
So the solutions to the second equation are:
[tex]\[ v = -7.5 \quad\text{and}\quad v = 5 \][/tex]
### Equation 3:
[tex]\[ \frac{x^2 - 10}{x - 1} = \frac{-14 - 5x}{x - 1} \][/tex]
Since the denominators are the same, we can equate the numerators directly:
[tex]\[ x^2 - 10 = -14 - 5x \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 + 5x - 10 + 14 = 0 \][/tex]
[tex]\[ x^2 + 5x + 4 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 4)(x + 1) = 0 \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
So the solutions to the third equation are:
[tex]\[ x = -4 \quad\text{and}\quad x = -1 \][/tex]
### Summary:
1. [tex]\( x = -5 \)[/tex]
2. [tex]\( v = -7.5 \)[/tex] and [tex]\( v = 5 \)[/tex]
3. [tex]\( x = -4 \)[/tex] and [tex]\( x = -1 \)[/tex]
Hence, the complete set of solutions is:
[tex]\[ x = -5 \][/tex]
[tex]\[ v = -7.5, v = 5 \][/tex]
[tex]\[ x = -4, x = -1 \][/tex]
### Equation 1:
[tex]\[ \frac{2 x}{x+1} + \frac{5}{2 x} = 2 \][/tex]
First, let's find a common denominator, which is [tex]\(2x(x+1)\)[/tex]:
[tex]\[ \frac{(2x)^2 + 5(x + 1)}{2x(x + 1)} = 2 \][/tex]
Simplify the numerator:
[tex]\[ \frac{4x^2 + 5x + 5}{2x(x + 1)} = 2 \][/tex]
Cross-multiply to clear the fraction:
[tex]\[ 4x^2 + 5x + 5 = 4x(x + 1) \][/tex]
Expand the right-hand side:
[tex]\[ 4x^2 + 5x + 5 = 4x^2 + 4x \][/tex]
Subtract [tex]\(4x^2\)[/tex] from both sides:
[tex]\[ 5x + 5 = 4x \][/tex]
Subtract [tex]\(4x\)[/tex] from both sides:
[tex]\[ x + 5 = 0 \][/tex]
Therefore, solving for [tex]\(x\)[/tex]:
[tex]\[ x = -5 \][/tex]
So the solution to the first equation is:
[tex]\[ x = -5 \][/tex]
### Equation 2:
[tex]\[ \frac{5}{v} + \frac{5}{v + 10} = \frac{4}{3} \][/tex]
Find a common denominator, which is [tex]\(v(v + 10)\)[/tex]:
[tex]\[ \frac{5(v + 10) + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]
Simplify the numerator:
[tex]\[ \frac{5v + 50 + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]
Combine like terms:
[tex]\[ \frac{10v + 50}{v(v + 10)} = \frac{4}{3} \][/tex]
Cross-multiply to clear the fraction:
[tex]\[ 3(10v + 50) = 4v(v + 10) \][/tex]
Expand both sides:
[tex]\[ 30v + 150 = 4v^2 + 40v \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 4v^2 + 40v - 30v - 150 = 0 \][/tex]
[tex]\[ 4v^2 + 10v - 150 = 0 \][/tex]
Divide the entire equation by 2 to simplify:
[tex]\[ 2v^2 + 5v - 75 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (2v + 15)(v - 5) = 0 \][/tex]
So, the solutions for [tex]\(v\)[/tex] are:
[tex]\[ 2v + 15 = 0 \implies v = -\frac{15}{2} = -7.5 \][/tex]
[tex]\[ v - 5 = 0 \implies v = 5 \][/tex]
So the solutions to the second equation are:
[tex]\[ v = -7.5 \quad\text{and}\quad v = 5 \][/tex]
### Equation 3:
[tex]\[ \frac{x^2 - 10}{x - 1} = \frac{-14 - 5x}{x - 1} \][/tex]
Since the denominators are the same, we can equate the numerators directly:
[tex]\[ x^2 - 10 = -14 - 5x \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 + 5x - 10 + 14 = 0 \][/tex]
[tex]\[ x^2 + 5x + 4 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 4)(x + 1) = 0 \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
So the solutions to the third equation are:
[tex]\[ x = -4 \quad\text{and}\quad x = -1 \][/tex]
### Summary:
1. [tex]\( x = -5 \)[/tex]
2. [tex]\( v = -7.5 \)[/tex] and [tex]\( v = 5 \)[/tex]
3. [tex]\( x = -4 \)[/tex] and [tex]\( x = -1 \)[/tex]
Hence, the complete set of solutions is:
[tex]\[ x = -5 \][/tex]
[tex]\[ v = -7.5, v = 5 \][/tex]
[tex]\[ x = -4, x = -1 \][/tex]
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