1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))


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Question Number 69995 by behi83417@gmail.com last updated on 29/Sep/19
$1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))$
$1 . {\begin{cases} \sqrt{x^{2} + y^{2}} + \sqrt{x^{2} - y^{2}} = a \\ \sqrt{x + y} + \sqrt{x - y} = b \end{cases} [a, b \in R^{+}]$ $2 . {\begin{cases} \sqrt{\sqrt{x} + y} + \sqrt{x + \sqrt{y}} = a \\ \sqrt{\sqrt{x} - y} + \sqrt{x - \sqrt{y}} = b \end{cases}$ $3 . {\begin{cases} x^{2} + y^{2} = (a - b) xy \\ x^{3} + y^{3} = ab (x - y) \end{cases}$ $4 . {\begin{cases} x + y + z = a \\ x^{2} + y^{2} + z^{2} = b \\ x^{3} + y^{3} + z^{3} = abxyz \end{cases}$ $5 . \sqrt{x^{2} + \sqrt{x}} + \sqrt{x^{2} - \sqrt{x}} = 2$ $6 . \sqrt{x^{2} + x + 1} + \sqrt{x^{2} + x - 1} + \sqrt{x^{2} - x - 1} = 1$ $7 . 16^{\sin^{2} x} + 16^{\cos^{2} x} = 10$ $8 . 2^{lnx} = x . \ln (x + \sqrt{2})$
	Answered by mr W last updated on 29/Sep/19

	$Q 7 :$ $l e t t = 16^{\sin^{2} x}$ $16^{\cos^{2} x} = 16^{1 - \sin^{2} x} = \frac{16}{16^{\sin^{2} x}} = \frac{16}{t}$ $t + \frac{16}{t} = 10$ $\Rightarrow t^{2} - 10 t + 16 = 0$ $(t - 2) (t - 8) = 0$ $\Rightarrow t = 2 o r 8$ $16^{\sin^{2} x} = 2$ $\Rightarrow \sin^{2} x = \frac{\ln 2}{\ln 16} = \frac{1}{4}$ $\Rightarrow \sin x = \pm \frac{1}{2}$ $\Rightarrow x = n π \pm \frac{π}{6}$ $16^{\sin^{2} x} = 8$ $\Rightarrow \sin^{2} x = \frac{\ln 8}{\ln 16} = \frac{3}{4}$ $\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}$ $\Rightarrow x = n π \pm \frac{π}{3}$
	Commented by behi83417@gmail.com last updated on 29/Sep/19

	$thanks in advance dear master .$ $You can't use 'macro parameter character #' in math mode$ $please try anothers! (ha ha excuse me)$
	Answered by mr W last updated on 29/Sep/19

	$Q 5 :$ $2 x^{2} + 2 \sqrt{x^{4} - x} = 4$ $\sqrt{x^{4} - x} = 2 - x^{2}$ $x^{4} - x = x^{4} - 4 x^{2} + 4$ $4 x^{2} - x - 4 = 0$ $\Rightarrow x = \frac{1 + \sqrt{65}}{8}$
	Answered by ajfour last updated on 29/Sep/19
	$(1) { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} let y=tx , ∣x∣=r (√(1+t^2 ))+(√(1−t^2 ))=(a/r) (√(1+t))+(√(1−t))=(b/(√r)) (√(1+t^2 ))+(√(1−t^2 ))=((2a)/b^2 )(1+(√(1−t^2 )) ) let t^2 =cos 2θ , ((2a)/b^2 )=μ (√2)cos θ+(√2)sin θ=(((2a)/b^2 ))(1+(√2)sin θ) cos θ+(1−μ)sin θ=(μ/(√2)) cos {φ−θ}=(μ/((√2)(√(1+(1−μ)^2 )))) φ=tan^(−1) (1−μ) =tan^(−1) (1−((2a)/b^2 )) t^2 =cos 2θ θ = φ−cos^(−1) (μ/((√2)(√(1+(1−μ)^2 )))) x=r = (a/((√(1+t^2 ))+(√(1−t^2 )))) y=tx .$
	$(1) {\begin{cases} \sqrt{x^{2} + y^{2}} + \sqrt{x^{2} - y^{2}} = a \\ \sqrt{x + y} + \sqrt{x - y} = b \end{cases}$ $l e t y = t x, ∣ x ∣= r$ $\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}} = \frac{a}{r}$ $\sqrt{1 + t} + \sqrt{1 - t} = \frac{b}{\sqrt{r}}$ $\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}} = \frac{2 a}{b^{2}} (1 + \sqrt{1 - t^{2}})$ $l e t t^{2} = \cos 2 θ, \frac{2 a}{b^{2}} = μ$ $\sqrt{2} \cos θ + \sqrt{2} \sin θ = (\frac{2 a}{b^{2}}) (1 + \sqrt{2} \sin θ)$ $\cos θ + (1 - μ) \sin θ = \frac{μ}{\sqrt{2}}$ $\cos {ϕ - θ} = \frac{μ}{\sqrt{2} \sqrt{1 + {(1 - μ)}^{2}}}$ $ϕ = \tan^{- 1} (1 - μ) = \tan^{- 1} (1 - \frac{2 a}{b^{2}})$ $t^{2} = \cos 2 θ$ $θ = ϕ - \cos^{- 1} \frac{μ}{\sqrt{2} \sqrt{1 + {(1 - μ)}^{2}}}$ $x = r = \frac{a}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}}$ $y = t x .$
	Commented by behi83417@gmail.com last updated on 30/Sep/19

	$thank you very much sir Ajfour .$ $you have creative mind, questions and$ $solutions . god bless you sir .$
	Answered by mr W last updated on 01/Oct/19

	$(3)$ $x + y = u$ $x y = v$ ${(x + y)}^{2} - 2 x y = (a - b) x y$ $u^{2} = (a - b + 2) v$ $x^{3} + y^{3} = (x + y) [{(x + y)}^{2} - 3 x y] = u (u^{2} - 3 v)$ $x - y = \sqrt{{(x + y)}^{2} - 4 x y} = \sqrt{u^{2} - 4 v}$ $u (u^{2} - 3 v) = a b \sqrt{u^{2} - 4 v}$ $u v (a - b - 1) = a b \sqrt{u^{2} - 4 v}$ $u^{2} v^{2} {(a - b - 1)}^{2} = a^{2} b^{2} (u^{2} - 4 v)$ $(a - b + 2) {(a - b - 1)}^{2} v^{2} = a^{2} b^{2} (a - b - 2)$ $\Rightarrow v = \frac{a b}{a - b - 1} \sqrt{\frac{a - b - 2}{a - b + 2}} = p s a y$ $\Rightarrow u = \pm \sqrt{\frac{a b \sqrt{(a - b + 2) (a - b - 2)}}{a - b - 1}} = \pm q s a y$ $x^{2} \pm q x + p = 0$ $x, y = \frac{1}{2} (\pm q \pm \sqrt{q^{2} - 4 p})$
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