Question

solve 1/a+b+x = 1/a+1/b+1/x x not equal to 0,-(a+b)

Answer 1

Answer:

Solution :

1/a+b+x=(1/a)+(1/b)+(1/x)

There's a trick in solving this question there is only one way to solve .

\begin{lgathered}\frac{1}{a + b + x} = \: \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \\ \frac{1}{ a+ b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \\. \frac{x - a - b - x}{ax + bx + {x}^{2} } = \frac{a + b}{ab} \\ \frac{ - (a + b)}{ax+bx+ {x}^{2} } = \frac{a + b}{ab} \\ \frac{ - 1}{ax + bx + {x}^{2} } = \frac{1}{ab} \\ - ab = ax + bx + {x}^{2} \\\end{lgathered}

a+b+x

1

=

a

1

+

b

1

+

x

1

a+b+x

1

−

x

1

=

a

1

+

b

1

.

ax+bx+x

2

x−a−b−x

=

ab

a+b

ax+bx+x

2

−(a+b)

=

ab

a+b

ax+bx+x

2

−1

=

ab

1

−ab=ax+bx+x

2

now after this we will solve the quadratic equation.

x^2+ax+bx+ab=0

x(x+a)+b(x+a)=0

(x+b)(x+a)=0

x= -a

x= -b

Step-by-step explanation:

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Answer 2

Correct Question :-

Solve 1/(a+b+x) = 1/a+1/b+1/x ,x ≠ 0 ,x ≠ -(a + b)

Solution :-

$\rm\:\dfrac{1}{(a+b+x)}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}$

$\rm\longrightarrow\:\dfrac{1}{(a+b+x)}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}$

$\rm\longrightarrow\:\dfrac{x-(a+b+x)}{x(a+b+x)}=\dfrac{b+a}{ab}$

$\rm\longrightarrow\:\dfrac{-(a+b)}{x(a+b+x)}=\dfrac{(a+b)}{ab}$

Dividing both side by (a + b), we get

$\rm\longrightarrow\:\dfrac{-1}{x(a+b+x)}=\dfrac{-1}{ab}$

By cross multiplication,

$\rm\longrightarrow\:x(a+b+x)=-ab$

$\rm\longrightarrow\:{x}^{2}+ax+bx+ab=0$

$\rm\longrightarrow\:x(x+a)+b(x+a)=0$

$\rm\longrightarrow\:(x+a)(x+b)=0$

Now,

$\rm\:x+a=0$

$\rm\longrightarrow\:x=0-a$

$\rm\longrightarrow\:x=-a$

Then,

$\rm\:x+b=0$

$\rm\longrightarrow\:x=0-b$

$\rm\longrightarrow\:x=-b$

Hence, -a and -b are the roots of the given equation,

More Information :-

Roots of a quadratic equations :-A real number α is called a root of the quadratic equation ax² + bx + c = 0 where a ≠ 0.