The value of x - 3 - x 2 - x - 6 - 2 x - 1 - 2 x 2 - 5 x - 3 - 2 x - 5 - x 2 - 5 x - 6 is-

The correct option is D 0 x 3 / x ^ 2 x 6 + 2x 1 / 2x ^ 2 +5x 3 2x+5 / x ^ 2 +5x+6 = x 3 /( x 3 ) ( x+2 ) #160; + 2x 1 /( x+3 ) ( 2x 1 ...

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Byju's Answer

Standard X

Mathematics

Quadratic Formula

The value of ...

Question

The value of $(\frac{x - 3}{x^{2} - x - 6} + \frac{2 x - 1}{2 x^{2} + 5 x - 3} - \frac{2 x + 5}{x^{2} + 5 x + 6})$ is:

\frac{x - 3}{x - 2}

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Solution

The correct option is D $0$
$\frac{x - 3}{x^{2} - x - 6} + \frac{2 x - 1}{{2 x}^{2} + 5 x - 3} - \frac{2 x + 5}{x^{2} + 5 x + 6} = \frac{x - 3}{(x - 3) (x + 2)} + \frac{2 x - 1}{(x + 3) (2 x - 1)} - \frac{2 x + 5}{(x + 2) (x + 3)} (x \neq 3, - 3, - 2, \frac{1}{2}) = \frac{1}{x + 2} + \frac{1}{x + 3} - \frac{(x + 2) + (x + 3)}{(x + 2) (x + 3)} = \frac{1}{x + 2} + \frac{1}{x + 3} - \frac{1}{x + 3} - \frac{1}{x + 2} = 0$
so option D is correct

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Similar questions

Q. If

x^{2} - x + 1 = 0

then the value of

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. If

x^{2} - x + 1 = 0

, then the value of

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

is equal to

Q. Add

(3 x^{2} - \frac{1}{5} x + \frac{7}{3}) + (- \frac{1}{4} x^{2} + \frac{1}{3} x + \frac{1}{3} x - \frac{1}{6}) + (- 2 x^{2} - \frac{1}{2} x + 5)

Find each of the following products.
(i) $(x - 4) (x - 4)$
(ii) $(2 x - 3 y) (2 x - 3 y)$
(iii) $(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y)$
(iv) $(x - \frac{3}{x}) (x - \frac{3}{x})$
(v) $(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9)$
(vi) $(\frac{1}{2} y^{2} - \frac{1}{3} y) (\frac{1}{2} y^{2} - \frac{1}{3} y)$

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The value of (x−3x2−x−6+2x−12x2+5x−3−2x+5x2+5x+6) is:

The correct option is D 0x−3x2−x−6+2x−12x2+5x−3−2x+5x2+5x+6=x−3(x−3)(x+2)+2x−1(x+3)(2x−1)−2x+5(x+2)(x+3)(x≠3,−3,−2,12)=1x+2+1x+3−(x+2)+(x+3)(x+2)(x+3)=1x+2+1x+3−1x+3−1x+2=0so option D is correct

The value of $(\frac{x - 3}{x^{2} - x - 6} + \frac{2 x - 1}{2 x^{2} + 5 x - 3} - \frac{2 x + 5}{x^{2} + 5 x + 6})$ is: