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Answer 1

$\dag\:\underline{\sf AnsWer :}$

$\longrightarrow\:\:\sf \dfrac{1}{x - 1} - \dfrac{1}{x } = \dfrac{1}{x + 3 } - \dfrac{1}{x + 4}$

$\longrightarrow\:\:\sf \dfrac{x - (x - 1)}{(x - 1)(x)} = \dfrac{x + 4 - (x + 3)}{(x + 3) (x + 4)}$

$\longrightarrow\:\:\sf \dfrac{x - x + 1}{ {x}^{2} - x} = \dfrac{x + 4 - x - 3}{ {x}^{2} + 4x + 3x + 12}$

$\longrightarrow\:\:\sf \dfrac{1}{ {x}^{2} - x} = \dfrac{4 - 3}{ {x}^{2} + 7x + 12}$

$\longrightarrow\:\:\sf \dfrac{1}{ {x}^{2} - x} = \dfrac{1}{ {x}^{2} + 7x + 12}$

$\longrightarrow\:\:\sf {x}^{2} + 7x + 12 =x^{2} - x$

$\longrightarrow\:\:\sf {x}^{2} - {x}^{2} + 12 =- x - 7x$

$\longrightarrow\:\:\sf 12 =- 8x$

$\longrightarrow\:\:\sf x = \dfrac{12}{ - 8}$

$\longrightarrow\:\:\sf x = - \dfrac{ 12}{8}$

$\longrightarrow\:\:\sf x = - \dfrac{6}{4}$

$\longrightarrow\:\:\sf x = - \dfrac{3}{2}$

$\longrightarrow\:\: \underline{ \boxed{\frak{ x = - 1.5 }}}$

Answer 2

$\large\sf\underline{Given}$

• $\sf\:\frac{1}{x-1}-\frac{1}{x}=\frac{1}{x+3}-\frac{1}{x+4}$

$\large\sf\underline{To\:find}$

• x = ?

$\large\sf\underline{Solution}$

$\sf➞\:\frac{1}{x-1}-\frac{1}{x}=\frac{1}{x+3}-\frac{1}{x+4}$

• Let's find the LCM

$\sf\:\frac{x-(x-1)}{x(x-1)}=\frac{x+4-(x+3)}{(x+3)(x+4)}$

$\sf\:\frac{x-x+1}{x^{2}-x}=\frac{x+4-x-3}{x^{2}+4x+3x+12}$

$\sf\:\frac{1}{x^{2}-x}=\frac{4-3}{x^{2}+7x+12}$

$\sf\:\frac{1}{x^{2}-x}=\frac{1}{x^{2}+7x+12}$

• Cross multiplying

$\sf\:x^{2}+7x+12=x^{2}-x$

$\sf\:x^{2}-x^{2}+7x+x+12=0$

$\sf\:7x+x+12=0$

$\sf\:8x+12=0$

$\sf\:8x=-12$

$\sf\:x=\frac{\cancel{-12}}{\cancel{8}}$

$\sf\:x=\frac{-3}{2}$

$\small{\underline{\boxed{\mathrm\pink{\:x=-1.5}}}}$

!! Hope it helps !!