Equation
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FacuFeri 06/03/2019 at 05:13
We have : \(x^2-180x+8102=\left(x^2-180x+8100\right)+2=\left(x-90\right)^2+2\ge2\forall x\left(1\right)\)
Applying the Bunhiacopxki inequality , we have :
\(\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le\left(1+1\right)\left(x-89+91-x\right)\)
\(\Rightarrow\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le2.2=4\)
\(\Rightarrow\sqrt{x-89}+\sqrt{91-x}\le2\left(2\right)\)
Because \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)
So ( 1 ) ; ( 2 ) ; we have : \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102=2\)
Equal sign occurs \(\Leftrightarrow\sqrt{x-89}=\sqrt{91-x};x-90=0\left(3\right)\)
We have : \(\sqrt{x-89}=\sqrt{91-x}\) \(\Leftrightarrow x-89=91-x\Leftrightarrow x=90\left(4\right)\)
( 3 ) ; ( 4 ) \(\Rightarrow x=90\) is the result of the equation
Lê Anh Duy selected this answer. -
Huy Toàn 8A (TL) 06/03/2019 at 13:14
Đk : \(89\le x\le91\)
Applying the Bunhiacopxki inequality
We have : VT \(=1.\sqrt{x-89}+1.\sqrt{91-x}\le\sqrt{\left(1+1\right)\left(x-89+91-x\right)=2}\)
=> VT \(\le2\)
And VP = \(x^2-2.x.90+90^2+2=\left(x-90\right)^2+2\ge2\)
=> VP \(\ge2\ge\) VT
The sign "=" occurs when and only when:
\(\left\{{}\begin{matrix}\sqrt{x-89}=\sqrt{91-x}\\x-90=0\end{matrix}\right.\) => \(x=90\)(satisfy)
The answer is 90
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FacuFeri 05/03/2019 at 05:10
Tran Anh ??? Le Quoc Tran Anh ???
Quoc Tran Anh Le Coordinator
05/08/2018 at 03:46
Lê Quốc Trần Anh Coordinator
11/07/2017 at 09:50
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The smallest and the biggest value of x are 1001 and 1111.The number of values of x is : (1111 - 1001) : 11 + 1 = 11
Lê Quốc Trần Anh selected this answer.
Dung Trần Thùy
19/03/2017 at 21:56
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mathlove 16/03/2017 at 14:00
With the condition \(x\ne0\), \(f\left(x^{-1}\right)=\dfrac{2x^{-1}+3}{2-3x^{-1}}=\dfrac{2+3x}{2x-3}\). The done equation become
\(\dfrac{2-3x}{2x+3}=\dfrac{2\left(2+3x\right)}{2x-3}\) .
With the additional condition \(x\ne\pm\dfrac{3}{2}\) , the above equation equivalent to
\(2\left(2+3x\right)\left(2x+3\right)=\left(2-3x\right)\left(2x-3\right)\)\(\Leftrightarrow18x^2-13x+18=0\)
This equation has no solution.
Selected by MathYouLike -
FA KAKALOTS 28/01/2018 at 22:11
With the condition x≠0, f(x−1)=2x−1+32−3x−1=2+3x2x−3
. The done equation become
2−3x2x+3=2(2+3x)2x−3
.
With the additional condition x≠±32
, the above equation equivalent to
2(2+3x)(2x+3)=(2−3x)(2x−3)
⇔18x2−13x+18=0
This equation has no solution.
Min Hoang moderators
16/03/2017 at 10:34
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mathlove 16/03/2017 at 11:33
Put \(t=x^2+18x-24\Leftrightarrow x^2+18x-24-t=0\Leftrightarrow x=-9\pm\sqrt{105+t},\left(t\ge-105\right)\), the equation become
\(\dfrac{1}{t+4}+\dfrac{1}{t}=\dfrac{-12}{t-5}\) (1)
With the condition \(t\notin\left\{0;-4;5\right\}\), \(\left(1\right)\Leftrightarrow7t^2+21t-10=0\Leftrightarrow t=\dfrac{-21\pm\sqrt{721}}{14}\)(both of these solutions are not smaller -105). So, the done equation have 4 solutions
\(x=-9\pm\pm\sqrt{\dfrac{1449+\sqrt{721}}{14}};x=-9\pm\sqrt{\dfrac{1449-\sqrt{721}}{14}}\) .
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FA KAKALOTS 28/01/2018 at 22:11
put t=x2+18x−24⇔x2+18x−24−t=0⇔x=−9±√105+t,(t≥−105)
, the equation become
1t+4+1t=−12t−5
(1)
With the condition t∉{0;−4;5}
, (1)⇔7t2+21t−10=0⇔t=−21±√72114
(both of these solutions are not smaller -105). So, the done equation have 4 solutions
x=−9±±√1449+√72114;x=−9±√1449−√72114
.
Min Hoang moderators
16/03/2017 at 10:16
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FA KAKALOTS 28/01/2018 at 22:11
x = c is a solution of the equation.
The equation is writed by x2−(a+b)x+c(a+b−c)=0
. By the theorem Viette, the rest solution is x=a+b−c .