Solve for x- i x-1-x-2- x-3-x-4 - 3 1-3- x - 2- 4- ii 1-x- 2-2x-3 - 1-x-2- x - 0- 3-2- 2- iii x - 1-x - 3- x - 0- iv 16-x-1 - 15-x-1- x - 0- - 1- v1-x-3 - 1-x-5 - 1-6- x - 3- - 5-

(i) 5, 5/2 (ii)1,3 (iii)3 ±√(5)/2 (iv) ± 4 (v) 9, 7 ...

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Byju's Answer

Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

Solve for x: ...

Question

Solve for x:
$(i) \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}; x \neq 2, 4 (i i) \frac{1}{x} + \frac{2}{2 x - 3} = \frac{1}{x - 2}, x \neq 0, \frac{3}{2}, 2 (i i i) x + \frac{1}{x} = 3, x \neq 0 (i v) \frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 (v) \frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5,$

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Solution

$(i) 5, \frac{5}{2} (i i) 1, 3 (i i i) \frac{3 \pm \sqrt{5}}{2} (i v) \pm 4 (v) - 9, 7$

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Q. Solve for

x

\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}; x \neq {3, 5}

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\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, \frac{- 3}{2}

Find the following integrals.
If $\frac{d}{d x} f (x) = 4 x^{3} - \frac{3}{x^{4}}$ such that f(2)=0. Then f(x)is
$(a) x^{4} + \frac{1}{x^{3}} - \frac{129}{8} (b) x^{3} + \frac{1}{x^{4}} + \frac{129}{8} (c) x^{4} + \frac{1}{x^{3}} + \frac{129}{8} (d) x^{3} + \frac{1}{x^{4}} - \frac{129}{8}$

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Q. Evaluate:
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(ii)

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}, \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}, 3 x + y \neq 0, 3 x - y \neq 0

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Solve for x: (i)x−1x−2+x−3x−4=313;x≠2,4(ii)1x+22x−3=1x−2,x≠0,32,2(iii)x+1x=3,x≠0(iv)16x−1=15x+1,x≠0,−1(v)1x−3−1x+5=16,x≠3,−5,

(i)5,52(ii)1,3(iii)3±√52(iv)±4(v)−9,7

$(i) 5, \frac{5}{2} (i i) 1, 3 (i i i) \frac{3 \pm \sqrt{5}}{2} (i v) \pm 4 (v) - 9, 7$