Equation

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

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

Tran Anh ??? Le Quoc Tran Anh ??? 

The smallest and the biggest value of x are 1001 and 1111.The number of values of x is : (1111 - 1001) : 11 + 1 = 11

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.

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.

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

                 \(x=-9\pm\pm\sqrt{\dfrac{1449+\sqrt{721}}{14}};x=-9\pm\sqrt{\dfrac{1449-\sqrt{721}}{14}}\)  .

​ ​

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

  .

x = c is a solution of the equation. 

The equation is writed by  \(x^2-\left(a+b\right)x+c\left(a+b-c\right)=0\) . By the theorem Viette, the rest solution is  \(x=a+b-c\) .

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 .