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Sagot :
Let's solve the equation step-by-step:
Given the equation:
[tex]\[ \frac{8x}{x+1} - 5 = \frac{8}{x^2 + x} \][/tex]
First, to clear the denominators, let's find a common denominator. Notice that:
[tex]\[ x^2 + x = x(x + 1) \][/tex]
Thus, the common denominator for all terms is [tex]\( x(x + 1) \)[/tex].
Rewriting each term with this common denominator, we get:
[tex]\[ \frac{8x(x)}{(x+1)x} - \frac{5(x^2 + x)}{x^2 + x} = \frac{8(x+1)}{x(x+1)} \][/tex]
Simplifying each term:
[tex]\[ \frac{8x^2}{x^2 + x} - \frac{5(x^2 + x)}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Combining the fractions on the left-hand side:
[tex]\[ \frac{8x^2 - 5(x^2 + x)}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Expanding and simplifying the numerator on the left-hand side:
[tex]\[ \frac{8x^2 - 5x^2 - 5x}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
So, it simplifies to:
[tex]\[ \frac{3x^2 - 5x}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Since the denominators are the same, we can equate the numerators:
[tex]\[ 3x^2 - 5x = 8 \][/tex]
Rearranging to form a standard quadratic equation:
[tex]\[ 3x^2 - 5x - 8 = 0 \][/tex]
To solve for [tex]\( x \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -8 \)[/tex]. Plugging these values in:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 96}}{6} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{121}}{6} \][/tex]
Since [tex]\(\sqrt{121} = 11\)[/tex], we have:
[tex]\[ x = \frac{5 \pm 11}{6} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3} \][/tex]
and
[tex]\[ x = \frac{5 - 11}{6} = \frac{-6}{6} = -1 \][/tex]
However, we must check if these values satisfy the original equation. Looking at the original equation:
[tex]\[ \frac{8x}{x+1} - 5 = \frac{8}{x^2 + x} \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{8(-1)}{-1+1} - 5 = \frac{8}{(-1)^2 + (-1)} \][/tex]
[tex]\[ \frac{-8}{0} - 5 = \frac{8}{0} \][/tex]
Both terms are undefined since division by zero is not possible, hence [tex]\( x = -1 \)[/tex] is an extraneous solution.
Thus, the only valid solution is:
[tex]\[ x = \frac{8}{3} \][/tex]
Given the equation:
[tex]\[ \frac{8x}{x+1} - 5 = \frac{8}{x^2 + x} \][/tex]
First, to clear the denominators, let's find a common denominator. Notice that:
[tex]\[ x^2 + x = x(x + 1) \][/tex]
Thus, the common denominator for all terms is [tex]\( x(x + 1) \)[/tex].
Rewriting each term with this common denominator, we get:
[tex]\[ \frac{8x(x)}{(x+1)x} - \frac{5(x^2 + x)}{x^2 + x} = \frac{8(x+1)}{x(x+1)} \][/tex]
Simplifying each term:
[tex]\[ \frac{8x^2}{x^2 + x} - \frac{5(x^2 + x)}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Combining the fractions on the left-hand side:
[tex]\[ \frac{8x^2 - 5(x^2 + x)}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Expanding and simplifying the numerator on the left-hand side:
[tex]\[ \frac{8x^2 - 5x^2 - 5x}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
So, it simplifies to:
[tex]\[ \frac{3x^2 - 5x}{x^2 + x} = \frac{8}{x^2 + x} \][/tex]
Since the denominators are the same, we can equate the numerators:
[tex]\[ 3x^2 - 5x = 8 \][/tex]
Rearranging to form a standard quadratic equation:
[tex]\[ 3x^2 - 5x - 8 = 0 \][/tex]
To solve for [tex]\( x \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -8 \)[/tex]. Plugging these values in:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 96}}{6} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{121}}{6} \][/tex]
Since [tex]\(\sqrt{121} = 11\)[/tex], we have:
[tex]\[ x = \frac{5 \pm 11}{6} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3} \][/tex]
and
[tex]\[ x = \frac{5 - 11}{6} = \frac{-6}{6} = -1 \][/tex]
However, we must check if these values satisfy the original equation. Looking at the original equation:
[tex]\[ \frac{8x}{x+1} - 5 = \frac{8}{x^2 + x} \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{8(-1)}{-1+1} - 5 = \frac{8}{(-1)^2 + (-1)} \][/tex]
[tex]\[ \frac{-8}{0} - 5 = \frac{8}{0} \][/tex]
Both terms are undefined since division by zero is not possible, hence [tex]\( x = -1 \)[/tex] is an extraneous solution.
Thus, the only valid solution is:
[tex]\[ x = \frac{8}{3} \][/tex]
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