((−♭o♭−)/(hans)) (1)(√(x−(√(x−(1/4))))) ≥ (1/4) (2)∣a^→ ∣ = 1, ∣b^→ ∣ = 2 , ∣c^→ ∣=3 , ∠(a^→ ,b^→ )=90° ∠(b^→ ,c^→ )=60° , ∠(a^→ ,c^→ )=120° , then ∣a^→ +b^→ −c^→ ∣=? (3) { ((x^(log _3 (y)) +y^(log _3 (x)) =18)),((log _3 (x)+log


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Question Number 109584 by bobhans last updated on 24/Aug/20
$((−♭o♭−)/(hans)) (1)(√(x−(√(x−(1/4))))) ≥ (1/4) (2)∣a^→ ∣ = 1, ∣b^→ ∣ = 2 , ∣c^→ ∣=3 , ∠(a^→ ,b^→ )=90° ∠(b^→ ,c^→ )=60° , ∠(a^→ ,c^→ )=120° , then ∣a^→ +b^→ −c^→ ∣=? (3) { ((x^(log _3 (y)) +y^(log _3 (x)) =18)),((log _3 (x)+log _3 (y)=3)) :} . Find the solution$
$\frac{- ♭ o ♭ -}{h a n s}$ $(1) \sqrt{x - \sqrt{x - \frac{1}{4}}} ⩾ \frac{1}{4}$ $(2) ∣ \vec{a} ∣ = 1, ∣ \vec{b} ∣ = 2, ∣ \vec{c} ∣= 3, ∠ (\vec{a}, \vec{b}) = 90 °$ $∠ (\vec{b}, \vec{c}) = 60 °, ∠ (\vec{a}, \vec{c}) = 120 °, t h e n$ $∣ \vec{a} + \vec{b} - \vec{c} ∣= ?$ $(3) {\begin{cases} x^{\log_{3} (y)} + y^{\log_{3} (x)} = 18 \\ \log_{3} (x) + \log_{3} (y) = 3 \end{cases} . F i n d t h e s o l u t i o n$
Commented by bemath last updated on 25/Aug/20
$•((♭ε)/(math))• (3) { ((y^(log _3 (x)) +y^(log _3 (x)) =18)),((log _3 (x)+log _3 (y) = 3)) :}→ { ((y^(log _3 (x)) =9)),((log _3 (x)+log _3 (y)=3)) :} log _3 (y)×log _3 (x)=2 ∧log _3 (x)+log _3 (y)=3 let { ((log _3 (x)=p)),((log _3 (y)=q)) :}→ { ((pq=2)),((p+q=3)) :} ⇔p(3−p)=2 →p^2 −3p+2=0 { ((p=1→q=2)),((p=2→q=1)) :}→ { ((x=3 ∧y=9)),((x=9 ∧y =3)) :}$
$∙ \frac{♭ ϵ}{m a t h} ∙$ $(3) {\begin{cases} y^{\log_{3} (x)} + y^{\log_{3} (x)} = 18 \\ \log_{3} (x) + \log_{3} (y) = 3 \end{cases} \to {\begin{cases} y^{\log_{3} (x)} = 9 \\ \log_{3} (x) + \log_{3} (y) = 3 \end{cases}$ $\log_{3} (y) \times \log_{3} (x) = 2 \land \log_{3} (x) + \log_{3} (y) = 3$ $l e t {\begin{cases} \log_{3} (x) = p \\ \log_{3} (y) = q \end{cases} \to {\begin{cases} p q = 2 \\ p + q = 3 \end{cases}$ $\Leftrightarrow p (3 - p) = 2 \to p^{2} - 3 p + 2 = 0$ ${\begin{cases} p = 1 \to q = 2 \\ p = 2 \to q = 1 \end{cases} \to {\begin{cases} x = 3 \land y = 9 \\ x = 9 \land y = 3 \end{cases}$
	Answered by bemath last updated on 24/Aug/20

	$\frac{↭ ♭ ϵ ↭}{Δ M a t h Δ}$ $(2) c o n s i d e r \vec{a} . \vec{a} = ∣ \vec{a} ∣^{2}$ $(\vec{a} + \vec{b} - \vec{c}) \circ (\vec{a} + \vec{b} - \vec{c}) = ∣ \vec{a} + \vec{b} - \vec{c} ∣^{2}$ $∣ \vec{a} ∣^{2} + 2 \vec{a} \circ \vec{b} - 2 \vec{a} \circ \vec{c} + ∣ \vec{b} ∣^{2} - 2 \vec{b} \circ \vec{c} + ∣ \vec{c} ∣=$ $∣ \vec{a} + \vec{b} - \vec{c} ∣^{2}$ $1 + 4 + 9 + 0 - 2 (1) (3) (- \frac{1}{2}) - 2 (2) (3) (\frac{1}{2}) =$ $∣ \vec{a} + \vec{b} - \vec{c} ∣^{2}$ $14 + 3 - 6 =∣ \vec{a} + \vec{b} - \vec{c} ∣^{2}$ $∴ ∣ \vec{a} + \vec{b} - \vec{c} ∣ = \sqrt{11}$
	Commented by bobhans last updated on 25/Aug/20

	$y e a h h . . .$
	Answered by john santu last updated on 24/Aug/20

	$\sqrt{x - \sqrt{x - \frac{1}{4}}} ⩾ \frac{1}{4}$ $\Rightarrow x - \sqrt{x - \frac{1}{4}} ⩾ \frac{1}{16}$ $\Rightarrow x - \frac{1}{16} ⩾ \sqrt{x - \frac{1}{4}}; x > \frac{1}{16} (1)$ $\Rightarrow x^{2} - \frac{x}{8} + \frac{1}{256} ⩾ x - \frac{1}{4}$ $\Rightarrow x^{2} - \frac{9 x}{8} + \frac{65}{256} ⩾ 0$ $\Rightarrow {(x - \frac{9}{16})}^{2} - \frac{81}{256} + \frac{65}{256} ⩾ 0$ $\Rightarrow {(x - \frac{9}{16})}^{2} - {(\frac{4}{16})}^{2} ⩾ 0$ $\Rightarrow (x - \frac{13}{16}) (x - \frac{5}{16}) ⩾ 0$ $\Rightarrow x ⩽ \frac{5}{16} \cup x ⩾ \frac{13}{16} (2)$ $\Rightarrow \sqrt{x - \frac{1}{4}} d e f i n e f o r x ⩾ \frac{4}{16} (3)$ $s o l u t i o n w e g e t f r o m (1) \cap (2) \cap (3)$ $x \in [\frac{1}{4}, \frac{5}{16}] \cup [\frac{13}{16}, \infty)$
	Commented by bobhans last updated on 25/Aug/20

	$y u h h u u y y$
	Answered by 1549442205PVT last updated on 24/Aug/20
	$Solve the inequality (1)(√(x−(√(x−(1/4))))) ≥ (1/4) We need must have the condition that x≥(1/4) in order to define the root.Under this condition squaring two sides of the inequality we get an equivalent inequality x−(√(x−(1/4))) ≥(1/(16))⇔x−(1/(16))≥(√(x−(1/4))) Since both sides are positive,squaring again we get ⇔x^2 −(1/8)x+(1/(256))≥x−(1/4) ⇔256x^2 −288x+65≥0(2) Δ′=144^2 −256.65=16^2 .(9^2 −65)=16^3 (√Δ) =64,x_(1,2) =((144±64)/(256))∈{((13)/(16));(5/(16))} ⇒(2)⇔x∈(−∞;(5/(16))]∪[((13)/(16));+∞) Combining to the condition x≥(1/4) we get roots of given inequality be x∈[(1/4);(5/(16))]∪[((13)/(16));+∞) 2)∣a^(→) +b^(→) −c^(→) ∣^2 =a^2 +b^2 +c^2 +2a^(→) .b^(→) −2a^(→) .c^(→) −2b^(→) .c^(→) =1^2 +2^2 +3^2 +2∣a∣.∣b∣cos90° −2∣a∣.∣c∣cos120°−2∣b∣.∣c∣cos60° =14+2.1.2×0−2×1×3(−0.5) −2.2.3.0.5=14+3−6=11 Hence,∣a^(→) +b^(→) −c^(→) ∣=(√(11))$
	$Solve the inequality$ $(1) \sqrt{x - \sqrt{x - \frac{1}{4}}} ⩾ \frac{1}{4}$ $We need must have the condition that$ $x ⩾ \frac{1}{4} in order to define the root . Under$ $this condition squaring two sides of$ $the inequality we get an equivalent$ $inequality$ $x - \sqrt{x - \frac{1}{4}} ⩾ \frac{1}{16} \Leftrightarrow x - \frac{1}{16} ⩾ \sqrt{x - \frac{1}{4}}$ $Since both sides are positive, squaring$ $again we get$ $\Leftrightarrow x^{2} - \frac{1}{8} x + \frac{1}{256} ⩾ x - \frac{1}{4}$ $\Leftrightarrow {256 x}^{2} - 288 x + 65 ⩾ 0 (2)$ $Δ^{'} = 144^{2} - 256.65 = 16^{2} . (9^{2} - 65) = 16^{3}$ $\sqrt{Δ} = 64, x_{1, 2} = \frac{144 \pm 64}{256} \in {\frac{13}{16}; \frac{5}{16}}$ $\Rightarrow (2) \Leftrightarrow x \in (- \infty; \frac{5}{16}] \cup [\frac{13}{16}; + \infty)$ $Combining to the condition x ⩾ \frac{1}{4}$ $we get roots of given inequality be$ $x \in [\frac{1}{4}; \frac{5}{16}] \cup [\frac{13}{16}; + \infty)$ $2) ∣ \vec{a} + \vec{b} - \vec{c} ∣^{2} = a^{2} + b^{2} + c^{2} + 2 \vec{a} . \vec{b} - 2 \vec{a} . \vec{c}$ $- 2 \vec{b} . \vec{c} = 1^{2} + 2^{2} + 3^{2} + 2 ∣ a ∣ . ∣ b ∣ \cos 90 °$ $- 2 ∣ a ∣ . ∣ c ∣ \cos 120 ° - 2 ∣ b ∣ . ∣ c ∣ \cos 60 °$ $= 14 + 2.1.2 \times 0 - 2 \times 1 \times 3 (- 0.5)$ $- 2.2.3.0.5 = 14 + 3 - 6 = 11$ $Hence, ∣ \vec{a} + \vec{b} - \vec{c} ∣= \sqrt{11}$
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