C-05-Q048mediumsingle_mcq

Solve $\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}$ :

a $x = -\dfrac{5}{2}$
b $x = \dfrac{5}{2}$
c $x = 0$
dno real solution

ব্যাখ্যা

Grouping the reciprocals into symmetric pairs and cross-multiplying collapses the equation to a single linear one. Because the four constants are symmetric, the root is their common axis, giving $x=-\dfrac{5}{2}$ .

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Equations in One Variable — সব প্রশ্ন দেখুন (170টি) →

Solve $\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}$ :

x = -\dfrac{5}{2}

x = \dfrac{5}{2}

x = 0

dno real solution

Solve 1x+1−1x+4=1x+2−1x+3\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}x+11​−x+41​=x+21​−x+31​:

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve 1x+1−1x+4=1x+2−1x+3\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}x+11​−x+41​=x+21​−x+31​:

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve $\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}$ :

Solve $\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}$ :