■(√(bemath))★ (1)If (√a) −(√b) = 20 , a,b∈R , find maximum value of a−5b ? (2)lim_(x→4) (((√x)−(√(3(√x)−2)))/(x^2 −16)) ? (3)∫ ((tan (ln x) tan (ln ((x/2))))/x) dx (4)((((√(3x−7)))^2 −2)/(x−3)) ≤ ((3−((√x))^2 )/(x−3))


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Question Number 110944 by bemath last updated on 01/Sep/20

$◼ \sqrt{bemath} ★$ $(1) If \sqrt{a} - \sqrt{b} = 20, a, b \in R, find maximum$ $value of a - 5 b ?$ $(2) \lim_{x \to 4} \frac{\sqrt{x} - \sqrt{3 \sqrt{x} - 2}}{x^{2} - 16} ?$ $(3) \int \frac{\tan (\ln x) \tan (\ln (\frac{x}{2}))}{x} dx$ $(4) \frac{{(\sqrt{3 x - 7})}^{2} - 2}{x - 3} ⩽ \frac{3 - {(\sqrt{x})}^{2}}{x - 3}$
Commented by bemath last updated on 01/Sep/20

Commented by 1549442205PVT last updated on 02/Sep/20

$If x = 0 then \sqrt{3 x - 7} = \sqrt{- 7} ?$
Commented by bemath last updated on 02/Sep/20

$\sqrt{{(3 x - 7)}^{2}} sir$
	Answered by john santu last updated on 01/Sep/20

	Commented by bemath last updated on 01/Sep/20

	$typo a - 5 b = 625 - 5 \times 25 = 625 - 125 = 500$
	Answered by Dwaipayan Shikari last updated on 01/Sep/20

	$2) \lim_{x \to 4} \frac{x - 3 \sqrt{x} + 2}{x - 4} . \frac{1}{\sqrt{x} + \sqrt{3 \sqrt{x} - 2}} . \frac{1}{x + 4}$ $\lim_{x \to 4} \frac{w^{2} - 3 w + 2}{w^{2} - 4} . \frac{1}{32} = \frac{(w - 1) (w - 2)}{(w - 2) (w + 2)} . \frac{1}{32} = \frac{\sqrt{x} - 1}{\sqrt{x} + 2} . \frac{1}{32} = \frac{1}{128}$
	Answered by 1549442205PVT last updated on 02/Sep/20
	$(√a)−(√b)=20⇒(√b)=(√a)−20.Hence, P=a−5b=a−5((√a)−20)^2 =a−5a+200(√a)−2000 =−4(a−50(√a)+625)+500 =−4((√a)−25)^2 +500≤500 The equality ocurrs if and only if { (((√a)−25=0)),(((√a)−(√b)=20)) :}⇔ { ((a=625)),((b=25)) :} Thus,P=a−5b has greatest value equal to 500 when (a,b)=(625,25)$
	$\sqrt{a} - \sqrt{b} = 20 \Rightarrow \sqrt{b} = \sqrt{a} - 20 . Hence,$ $P = a - 5 b = a - 5 {(\sqrt{a} - 20)}^{2}$ $= a - 5 a + 200 \sqrt{a} - 2000$ $= - 4 (a - 50 \sqrt{a} + 625) + 500$ $= - 4 {(\sqrt{a} - 25)}^{2} + 500 ⩽ 500$ $The equality ocurrs if and only if$ ${\begin{cases} \sqrt{a} - 25 = 0 \\ \sqrt{a} - \sqrt{b} = 20 \end{cases} \Leftrightarrow {\begin{cases} a = 625 \\ b = 25 \end{cases}$ $Thus, P = a - 5 b has greatest value$ $equal to 500 when (a, b) = (625, 25)$
	Answered by 1549442205PVT last updated on 02/Sep/20

	$2) Set x - 4 = t (x \to 4) \sim (t \to 0), x = t + 4$ $(2) \lim \frac{\sqrt{x} - \sqrt{3 \sqrt{x} - 2}}{x^{2} - 16} = \lim_{t \to 0} \frac{\sqrt{t + 4} - \sqrt{3 \sqrt{t + 4} - 2}}{{(t + 4)}^{2} - 16}$ $= \lim_{t \to 0} \frac{t + 4 - (3 \sqrt{t + 4} - 2)}{[\sqrt{t + 4} + \sqrt{3 \sqrt{t + 4} - 2]} (t^{2} + 8 t)}$ $= \lim_{t \to 0} \frac{t + 6 - 3 \sqrt{t + 4}}{[\sqrt{t + 4} + \sqrt{3 \sqrt{t + 4} - 2]} (t^{2} + 8 t)}$ $= \lim_{t \to 0} \frac{{(t + 6)}^{2} - {(3 \sqrt{t + 4})}^{2}}{[\sqrt{t + 4} + \sqrt{3 \sqrt{t + 4} - 2]} (t^{2} + 8 t) (t + 6 + 3 \sqrt{t + 4})}$ $= \lim_{t \to 0} \frac{t^{2} + 12 t + 36 - 9 t - 36}{[\sqrt{t + 4} + \sqrt{3 \sqrt{t + 4} - 2]} (t^{2} + 8 t) (t + 6 + 3 \sqrt{t + 4})}$ $= \lim_{t \to 0} \frac{t + 3}{[\sqrt{t + 4} + \sqrt{3 \sqrt{t + 4} - 2]} (t + 6 + 3 \sqrt{t + 4}) (t + 8)}$ $= \frac{3}{(2 + 2) (6 + 6) . 8} = \frac{1}{128}$
	Answered by 1549442205PVT last updated on 02/Sep/20
	$(4)((((√(3x−7)))^2 −2)/(x−3)) ≤ ((3−((√x))^2 )/(x−3)) ⇔((3x−7−2)/(x−3))≤((3−x)/(x−3)) (1).Under the condition { ((x≥(7/3))),((x≠3)) :} (1)⇔((3x−9)/(x−3))≤−1⇔((x−3)/(x−3))≤−1⇔1≤−1 the inequality has no roots$
	$(4) \frac{{(\sqrt{3 x - 7})}^{2} - 2}{x - 3} ⩽ \frac{3 - {(\sqrt{x})}^{2}}{x - 3}$ $\Leftrightarrow \frac{3 x - 7 - 2}{x - 3} ⩽ \frac{3 - x}{x - 3} (1) . Under the$ $condition {\begin{cases} x ⩾ \frac{7}{3} \\ x \neq 3 \end{cases}$ $(1) \Leftrightarrow \frac{3 x - 9}{x - 3} ⩽ - 1 \Leftrightarrow \frac{x - 3}{x - 3} ⩽ - 1 \Leftrightarrow 1 ⩽ - 1$ $the inequality has no roots$
	Commented by bemath last updated on 02/Sep/20

	$wrong sir . if \sqrt{{(3 x - 7)}^{2}} = ∣ 3 x - 7 ∣$ $not \sqrt{{(3 x - 7)}^{2}} = 3 x - 7$
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