1. x^2 −1+((x^4 −x^2 ))^(1/3) =10x 2. (√(x−(1/x)))+(√(1−(1/x)))=x 3. (√(2x−(8/x)))+2(√(1−(2/x)))≥x 4. (√(x^2 +x))+(√(x+2))≥(√(3(x^2 −2x+2))) 5. ((√(x+5))−(√(x−3)))(1+(√(x^2 +2x−15)))≥8 6. x^2 +3x+1=(x+1)(√(x^2 +1)) 7. 2x^2 +3x+7=(x+5)(√(2x^2 +1))


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Question Number 201491 by tri26112004 last updated on 07/Dec/23

$1 . x^{2} - 1 + \sqrt[3]{x^{4} - x^{2}} = 10 x$ $2 . \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ $3 . \sqrt{2 x - \frac{8}{x}} + 2 \sqrt{1 - \frac{2}{x}} ⩾ x$ $4 . \sqrt{x^{2} + x} + \sqrt{x + 2} ⩾ \sqrt{3 (x^{2} - 2 x + 2)}$ $5 . (\sqrt{x + 5} - \sqrt{x - 3}) (1 + \sqrt{x^{2} + 2 x - 15}) ⩾ 8$ $6 . x^{2} + 3 x + 1 = (x + 1) \sqrt{x^{2} + 1}$ $7 . 2 x^{2} + 3 x + 7 = (x + 5) \sqrt{2 x^{2} + 1}$
Commented by Rasheed.Sindhi last updated on 08/Dec/23

$You can't use 'macro parameter character #' in math mode$
	Answered by Rasheed.Sindhi last updated on 08/Dec/23
	$2. (√(x−(1/x))) +(√(1−(1/x)))=x Let (√(x−(1/x))) =a , (√(1−(1/x))) =b a+b=x ∧ a^2 −b^2 =x−1 ⇒a−b=((a^2 −b^2 )/(a+b))=((x−1)/x) { (( a+b=x)),((a−b=((x−1)/x)=1−(1/x))) :} ⇒ { ((2a=x+((x−1)/x)=x−(1/x)+1=a^2 +1)),((2b=x−((x−1)/x)=x+(1/x)−1=x−(1−(1/x))=x−b^2 )) :} ⇒ { ((a^2 −2a+1=0⇒(a−1)^2 =0⇒a=1⇒b^2 +b=a=1)),((b^2 +2b=x)) :} ⇒(√(x−(1/x))) =1⇒x−(1/x)=1⇒x^2 −x−1=0 ⇒x_(≥0) =((1+(√(1+4)))/2)=((1+(√5) )/2)$
	$2 . \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ $L e t \sqrt{x - \frac{1}{x}} = a, \sqrt{1 - \frac{1}{x}} = b$ $a + b = x \land a^{2} - b^{2} = x - 1$ $\Rightarrow a - b = \frac{a^{2} - b^{2}}{a + b} = \frac{x - 1}{x}$ ${\begin{cases} a + b = x \\ a - b = \frac{x - 1}{x} = 1 - \frac{1}{x} \end{cases}$ $\Rightarrow {\begin{cases} 2 a = x + \frac{x - 1}{x} = x - \frac{1}{x} + 1 = a^{2} + 1 \\ 2 b = x - \frac{x - 1}{x} = x + \frac{1}{x} - 1 = x - (1 - \frac{1}{x}) = x - b^{2} \end{cases}$ $\Rightarrow {\begin{cases} a^{2} - 2 a + 1 = 0 \Rightarrow {(a - 1)}^{2} = 0 \Rightarrow a = 1 \Rightarrow b^{2} + b = a = 1 \\ b^{2} + 2 b = x \end{cases}$ $\Rightarrow \sqrt{x - \frac{1}{x}} = 1 \Rightarrow x - \frac{1}{x} = 1 \Rightarrow x^{2} - x - 1 = 0$ $\Rightarrow x_{⩾ 0} = \frac{1 + \sqrt{1 + 4}}{2} = \frac{1 + \sqrt{5}}{2}$
	Answered by esmaeil last updated on 07/Dec/23

	$1 . x^{2} - 1 + x \sqrt[3]{x - \frac{1}{x}} = 10 x \Rightarrow$ $x - \frac{1}{x} + \sqrt[3]{x - \frac{1}{x}} = 10$ $x - \frac{1}{x} = p^{3} \Rightarrow p^{3} + p = 10 \Rightarrow$ $p^{3} - 8 + (p - 2) = 0$ $(p - 2) (p^{2} + 2 p + 5) = 0 \Rightarrow p = 2$ $\Rightarrow x^{2} - 8 x - 1 = 0 \Rightarrow x = 4 \pm \sqrt{17}$
	Answered by ajfour last updated on 07/Dec/23

	$★ x^{2} - 1 + \sqrt[3]{x^{4} - x^{2}} = 10 x$ $x - \frac{1}{x} = p$ $p + \sqrt[3]{p} = 10$ ${(10 - p)}^{3} = p$ $s a y 10 - p = t$ $t^{3} + t = 10$ $t = 2$ $p = 8 = x - \frac{1}{x}$ $x^{2} - 8 x - 1 = 0$ $x = 4 \pm \sqrt{17}$
	Answered by ajfour last updated on 07/Dec/23

	$★ ★ \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ $\sqrt{x - z} + \sqrt{1 - z} = x$ $1 - z = \sqrt{x - z} - \sqrt{1 - z}$ $\Rightarrow 1 - z + x = 2 \sqrt{x - z}$ $s a y \sqrt{x - z} = p$ $1 + p^{2} = 2 p$ $p = 1 \Rightarrow x - \frac{1}{x} = 1$ $x^{2} - x - 1 = 0$ $x = \frac{1 \pm \sqrt{5}}{2}$
	Answered by Calculusboy last updated on 08/Dec/23

	$S o l u t i o n :$ $6) s q u a r e b o t h s i d e s$ ${(x^{2} + 3 x + 1)}^{2} = {[(x + 1) \sqrt{x^{2} + 1}]}^{2}$ ${(x^{2} + 3 x)}^{2} + 2 (x^{2} + 3 x) + 1 = {(x + 1)}^{2} (x^{2} + 1)$ $x^{4} + 6 x^{3} + 9 x^{2} + 2 x^{2} + 6 x + 1 = (x^{2} + 2 x + 1) (x^{2} + 1)$ $x^{4} + 6 x^{3} + 11 x^{2} + 6 x + 1 = x^{4} + 2 x^{3} + x^{2} + x^{2} + 2 x + 1$ $x^{4} - x^{4} + 6 x^{3} - 2 x^{3} + 11 x^{2} - 2 x^{2} + 6 x - 2 x + 1 - 1 = 0$ $4 x^{3} + 9 x^{2} + 4 x = 0 (d i v i d e t h r o u g h b y x)$ $4 x^{2} + 9 x + 4 = 0 \Leftrightarrow b y u s i n g x = \frac{- b \underline{+} \sqrt{b^{2} - 4 a c}}{2 a}$ $a = 4, b = 9 a n d c = 4$ $x = \frac{- 9 \underline{+} \sqrt{9^{2} - 4 \times 4 \times 4}}{2 \times 4} \Leftrightarrow x = \frac{- 9 \underline{+} \sqrt{81 - 64}}{8}$ $x = \frac{- 9 \underline{+} \sqrt{17}}{8}$ $x_{1} = \frac{- 9 + \sqrt{17}}{8} Λ x_{2} = \frac{- 9 - \sqrt{17}}{8}$
	Answered by Calculusboy last updated on 08/Dec/23

	$S o l u t i o n :$ $7) s q u a r e b o t h s i d e s$ ${(2 x^{2} + 3 x + 7)}^{2} = {[(x + 5) \sqrt{2 x^{2} + 1}]}^{2}$ ${(2 x^{2} + 3 x)}^{2} + 14 (2 x^{2} + 3 x) + 49 = {(x + 5)}^{2} (2 x^{2} + 1)$ $4 x^{4} + 12 x^{3} + 9 x^{2} + 28 x^{2} + 42 x + 49 = (x^{2} + 10 x + 25) (2 x^{2} + 1)$ $4 x^{4} + 12 x^{3} + 37 x^{2} + 42 x + 49 = 2 x^{4} + 20 x^{3} + 50 x^{2} + x^{2} + 10 x + 25$ $4 x^{4} - 2 x^{4} + 12 x^{3} - 20 x^{3} + 37 x^{2} - 51 x^{2} + 42 x - 10 x + 49 - 25 = 0$ $2 x^{4} - 8 x^{3} - 14 x^{2} + 32 x + 24 = 0 (d i v i d e t h r o u g h b y 2)$ $x^{4} - 4 x^{3} - 7 x^{2} + 16 x + 12 = 0 (b y u s i n g t r y a n d e r r o r)$ $l e t x = - 2 \Rightarrow x + 2 = 0$ ${(- 2)}^{4} - 4 {(- 2)}^{3} - 7 {(- 2)}^{2} + 16 (- 2) + 12 = 0 \Leftrightarrow 0 = 0$ $(x + 2) [x^{3} - 6 x^{2} + 5 x + 6] = 0$ $x_{1} = - 2$ $x^{3} - 6 x^{2} + 5 x + 6 = 0$ $l e t x = 2 \Leftrightarrow x - 2 = 0$ ${(2)}^{3} - 6 {(2)}^{2} + 5 (2) + 6 = 0 \Leftrightarrow 0 = 0$ $(x - 2) [x^{2} - 4 x - 3] = 0$ $x - 2 = 0 \Rightarrow x_{2} = 2$ $x^{2} - 4 x - 3 = 0 \Leftrightarrow x = \frac{- b \underline{+} \sqrt{b^{2} - 4 a c}}{2 a}$ $a = 1, b = - 4 a n d c = - 3$ $x = \frac{- (- 4) \underline{+} \sqrt{{(- 4)}^{2} - 4 \times 1 \times (- 3)}}{2 \times 1}$ $x = \frac{4 \underline{+} \sqrt{16 + 12}}{2} \Leftrightarrow x = \frac{4 \underline{+} \sqrt{28}}{2} \Leftrightarrow x = \frac{4 \underline{+} 2 \sqrt{7}}{2} = 2 \underline{+} \sqrt{7}$ $x_{3} = \frac{4 + 2 \sqrt{7}}{2} = 2 + \sqrt{7} Λ x_{4} = \frac{4 - 2 \sqrt{7}}{2} = 2 - \sqrt{7}$ $∴ t h e v a l u e f o r (x) = (- 2, 2, 2 + \sqrt{7}, 2 - \sqrt{7})$
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