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Question Number 97236 by bemath last updated on 07/Jun/20

Commented by mr W last updated on 07/Jun/20

$i f y o u p u t s e v e r a l q u e s t i o n s i n o n e$ $p o s t, p l e a s e n u m b e r t h e q u e s t i o n s!$
Commented by bemath last updated on 07/Jun/20

$hahaha . . . sorry sir$
Commented by bemath last updated on 07/Jun/20

$thank you for all$
	Answered by john santu last updated on 07/Jun/20

	Commented by mahdi last updated on 07/Jun/20

	$in case (2) \Rightarrow x + y = - 2 \land xy = - 1$ $but xy = (- 1 - \sqrt{3}) (- 1 + \sqrt{3}) = 1 - 3 = 2 ? ? ?$
	Answered by mahdi last updated on 19/Jun/20
	$1− { ((x^2 −xy+y^2 =7)),((x−xy+y=−1)) :}⇒^+ x^2 −2xy+y^2 +x+y=6 ⇒(x−y)^2 +(x+y)=6 [I] { ((x^2 −xy+y^2 =7)),((−3x+3xy−3y=3)) :}⇒^+ x^2 +2xy+y^2 −3(x+y)=10 ⇒(x+y)^2 −3(x+y)=10 [II] ⇒^(II) { (((x+y)=5⇒^I { (((x−y)=1 ⇒x=3,y=2)),(((x−y)=−1⇒x=2,y=3)) :})),(((x+y)=−2⇒^I { (((x−y)=2(√2)⇒x=(√2)−1,y=−(√2)−1)),(((x−y)=−2(√2)⇒x=−(√2)−1,y=(√2)−1)) :})) :} {(2,3),(3,2),((√2)−1,−(√2)−1),(−(√2)−1,(√2)−1)}⇒ans1 2− { (( D_f f(x) R_f )),((x<1 9−3x (6,+∞))),((1≤x≤3 7−x [4,6])),((3<x<5 x+1 (4,6) )),((5≤x 3x−9 [6,+∞))) :} min=4 3−((log((a/b)))/(log(b)))=((log(a)−log(b))/(log(b)))=((log(a))/(log(b)))−1 =1000−1=999 4−((3+x)/(4+x))=((6+x)/(8+x))⇒(3+x)(8+x)=(6+x)(4+x)⇒ x^2 +11x+24=x^2 +10x+24⇒x=0 5−S=((p(p−a)(p−b)(p−c)))^(1/�) (p=((a+b+c)/2)) (heron′L) S≈90$
	$1 - {\begin{cases} x^{2} - xy + y^{2} = 7 \\ x - xy + y = - 1 \end{cases} \overset{+}{\Rightarrow} x^{2} - 2 xy + y^{2} + x + y = 6$ $\Rightarrow {(x - y)}^{2} + (x + y) = 6 [I]$ ${\begin{cases} x^{2} - xy + y^{2} = 7 \\ - 3 x + 3 xy - 3 y = 3 \end{cases} \overset{+}{\Rightarrow} x^{2} + 2 xy + y^{2} - 3 (x + y) = 10$ $\Rightarrow {(x + y)}^{2} - 3 (x + y) = 10 [II]$ $\overset{II}{\Rightarrow} {\begin{cases} (x + y) = 5 \overset{I}{\Rightarrow} {\begin{cases} (x - y) = 1 \Rightarrow x = 3, y = 2 \\ (x - y) = - 1 \Rightarrow x = 2, y = 3 \end{cases} \\ (x + y) = - 2 \overset{I}{\Rightarrow} {\begin{cases} (x - y) = 2 \sqrt{2} \Rightarrow x = \sqrt{2} - 1, y = - \sqrt{2} - 1 \\ (x - y) = - 2 \sqrt{2} \Rightarrow x = - \sqrt{2} - 1, y = \sqrt{2} - 1 \end{cases} \end{cases}$ ${(2, 3), (3, 2), (\sqrt{2} - 1, - \sqrt{2} - 1), (- \sqrt{2} - 1, \sqrt{2} - 1)} \Rightarrow ans 1$ $2 - {\begin{cases} D_{f} f (x) R_{f} \\ x < 1 9 - 3 x (6, + \infty) \\ 1 ⩽ x ⩽ 3 7 - x [4, 6] \\ 3 < x < 5 x + 1 (4, 6) \\ 5 ⩽ x 3 x - 9 [6, + \infty) \end{cases}$ $\min = 4$ $3 - \frac{\log (\frac{a}{b})}{\log (b)} = \frac{\log (a) - \log (b)}{\log (b)} = \frac{\log (a)}{\log (b)} - 1$ $= 1000 - 1 = 999$ $4 - \frac{3 + x}{4 + x} = \frac{6 + x}{8 + x} \Rightarrow (3 + x) (8 + x) = (6 + x) (4 + x) \Rightarrow$ $x^{2} + 11 x + 24 = x^{2} + 10 x + 24 \Rightarrow x = 0$ $5 - S = \sqrt[]{p (p - a) (p - b) (p - c)} (p = \frac{a + b + c}{2}) ({heron}^{'} L)$ $S \approx 90$
	Commented by bemath last updated on 08/Jun/20

	$for eq (4) . how if x = 0 ?$ $\frac{3 + 0}{4 + 0} = \frac{6 + 0}{8 + 0} ? (it true)$
	Commented by bemath last updated on 08/Jun/20

	$if x = - \frac{14}{3} \Leftrightarrow \frac{3 + (- \frac{14}{3})}{4 + (- \frac{14}{3})} = \frac{9 - 14}{12 - 14}$ $= \frac{- 5}{- 2} = \frac{5}{2}$ $\Leftrightarrow \frac{6 + (- \frac{14}{3})}{8 + (- \frac{14}{3})} = \frac{18 - 14}{24 - 14} = \frac{4}{10} = \frac{2}{5}$ $clear \frac{5}{2} \neq \frac{2}{5} .$ $so x = - \frac{14}{3} not solution$
	Commented by bemath last updated on 08/Jun/20

	$the area triangle = 150$
	Commented by mahdi last updated on 19/Jun/20

	$thanks bemath for giude :)$ $but Area of Δ not 150 . . . ?$
	Answered by mr W last updated on 07/Jun/20

	Commented by PRITHWISH SEN 2 last updated on 07/Jun/20

	$Let A_{1}, A_{2}, A_{3} are the altitudes$ $then A = \frac{\frac{1}{A_{1}} + \frac{1}{A_{2}} + \frac{1}{A_{3}}}{2} = \frac{\frac{1}{12} + \frac{1}{15} + \frac{1}{20}}{2} = \frac{1}{10}$ $then area =$ $\frac{1}{4 \sqrt{A (A - \frac{1}{A_{1}}) (A - \frac{1}{A_{2}}) (A - \frac{1}{A_{3}})}}$ $= \frac{1}{4 \sqrt{\frac{1}{10} (\frac{1}{10} - \frac{1}{12}) (\frac{1}{10} - \frac{1}{15}) (\frac{1}{10} - \frac{1}{20})}} = 150$
	Commented by mr W last updated on 07/Jun/20

	$s a y t h e s i d e s o f t r i a n g l e a r e a, b, c a n d$ $c o r r e s p o n d i n g a l t i t u d e s a r e h_{a}, h_{b}, h_{c} .$ $s a y t h e a r e a o f t h e t r i a n g l e i s A .$ $w e h a v e$ $A = \frac{{a h}_{a}}{2} = \frac{{b h}_{b}}{2} = \frac{{c h}_{c}}{2}$ $\Rightarrow a = \frac{2 A}{h_{a}}, b = \frac{2 A}{h_{b}}, c = \frac{2 A}{h_{c}}$ $w i t h s = \frac{a + b + c}{2} = A (\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}})$ $w e a l s o h a v e$ $A = \sqrt{s (s - a) (s - b) (s - c)}$ $i . e .$ $A = \sqrt{A^{4} (\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (- \frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})}$ $\Rightarrow \frac{1}{A} = \sqrt{(\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (- \frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})}$ $o r$ $\Rightarrow A = \frac{1}{\sqrt{(\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (- \frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})}}$ $w i t h h_{a} = 12, h_{b} = 15, h_{c} = 20 w e g e t$ $A = \frac{1}{\sqrt{(\frac{1}{12} + \frac{1}{15} + \frac{1}{20}) (- \frac{1}{12} + \frac{1}{15} + \frac{1}{20}) (\frac{1}{12} - \frac{1}{15} + \frac{1}{20}) (\frac{1}{12} + \frac{1}{15} - \frac{1}{20})}}$ $= 150$
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