Đặt \(A=\dfrac{x^2-10x+25}{x^2-5}\)

ĐK : \(x^2-5\ne0\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne\sqrt{5}\\x\ne-\sqrt{5}\end{matrix}\right.\)

\(A=0\\ \Leftrightarrow\dfrac{x^2-10x+25}{x^2-5}=0\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TM\right)\)

Vậy x =5 thì A =0

 \(A=2018-\left|x-7\right|-\left|y+2\right|\)

Ta có: \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\\left|y+2\right|\ge0\forall y\end{cases}}\Rightarrow2018-\left|x-7\right|-\left|y+2\right|\le2018\)

\(A=2018\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}}\)

Vậy \(A_{m\text{ax}}=2018\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)

Tham khảo~

1)\(x^2=25\Rightarrow\left|x\right|=5\Rightarrow\int^{x=5}_{x=-5}\)
2)\(\left(x-2005\right)^2\ge0\Rightarrow\left(x-2005\right)^2+4\ge4\)
Dấu "=" xảy ra <=> x=2005
tick nhé

x2 = 25

x² = (-5)² = 5²

x thuộc {-5 ; 5}

(x - 2005)² + 4 có GTNN

(x - 2005)² + 4 \(\ge\) 4

Vậy GTNN (x-2005)2 + 4 = 4

Khi x -2005=  0 => x = 2005

1. x = -5 hoặc x = 5
2. GTNN = 4

B=x^4+2x^3-3x^2+2x-x^4-2x^3+3x^2-3x+x-12

=-12

\(B=x\left(x^3+2x^2-3x+2\right)-\left(x^2+2x\right)x^2+3x\left(x-1\right)+x-12\)

\(=x^4+2x^3-3x^2+2x-x^4-2x^3+3x^2-3x+x-12\)

\(=\left(x^4-x^4\right)+\left(2x^3-2x^3\right)+\left(-3x^2+3x^2\right)+\left(2x-3x+x\right)-12\)

\(=0+0+0+0-12\)

\(=-12\)

|x|-a=a=> |x|=a+a=0

 Vì |x|=0=>x=0 ^-^