Advertisements
Advertisements
प्रश्न
Solve the following equation :
`1/(("x" - 1)(x - 2)) + 1/(("x" - 2)("x" - 3)) + 1/(("x" - 3)("x" -4)) = 1/6`
Advertisements
उत्तर
`1/(("x" - 1)(x - 2)) + 1/(("x" - 2)("x" - 3)) + 1/(("x" - 3)("x" -4)) = 1/6`
`(("x" -3)("x" - 4) + ("x" - 1)("x" - 4) + ("x" - 1)("x" - 2))/(("x" - 1)("x" - 2)("x" - 3)("x" - 4)) = 1/6`
`("x"^2 - 3"x" - 4"x" + 12 + "x"^2 - "x" - 4"x" + 4 + "x"^2 - "x" - 2"x" + 2)/(("x" - 1)("x" - 2)("x" - 3)("x" - 4)) = 1/6`
`(3"x"^2 - 15 "x" + 18)/(("x" - 1)("x" - 2)("x" - 3)("x" - 4)) = 1/6`
`(3("x"^2 - 5"x" + 6))/(("x" - 1)("x" - 2)("x" - 3)("x" - 4)) = 1/6`
`(3("x" - 3)("x" - 2))/(("x" - 1)("x" - 2)("x" - 3)("x" - 4)) = 1/6`
`3/(("x" - 1)("x" - 4)) = 1/6`
x2 - 5x+ 4= 18
x2 - 5x - 14= 0
x2 + 2x - 7x -14= 0
x( x+ 2) - 7(x + 2)= 0
(x + 2)(x - 7) = 0
x = -2, x = 7
APPEARS IN
संबंधित प्रश्न
A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
The sum of natural number and its positive square root is 132. Find the number.
Solve the following quadratic equations by factorization: \[\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}; x \neq 5, 7\]
Solve the following equation: `("x" + 3)/("x" - 2) - (1 - "x")/"x" = 17/4`
The sum of the square of 2 positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.
If `sqrt (2/3)` is a solution of equation 3x2 + mx + 2 = 0, find the value of m.
Solve equation using factorisation method:
`6/x = 1 + x`
The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by `(4)/(35)`. Find the fraction.
Solve the following equation by factorisation :
`sqrt(x + 15) = x + 3`
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.
