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-

Question Number 109584 by bobhans last updated on 24/Aug/20

\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

∙ \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 \dots

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) \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}