Với  x ≥ 0 , x ≠ 25  thì  B = 3 x + 5 + 20 − 2 x x − 15 = 3 x + 5 + 20 − 2 x x + 5 x − 5

= 3 x − 5 + 20 − 2 x x + 5 x − 5 = 3 x − 15 + 20 − 2 x x + 5 x − 5 = x + 5 x + 5 x − 5 = 1 x − 5

 (điều phải chứng minh)

\(A=x+\left\{\left(x+5\right)-\left[\left(5-x\right)-\left(-x-3\right)\right]\right\}\)

\(=x+\left\{\left(x+5\right)-\left[5-x+x+3\right]\right\}\)

\(=x+\left\{\left(x+5\right)-\left(5+3\right)\right\}\)

\(=x+\left\{\left(x+5\right)-8\right\}\)

\(=x+\left\{x+5-8\right\}=x+\left\{x-3\right\}\)

\(=x+x-3=2x-3\)

\(B=x.\left\{\left[-x-2-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right]\right\}\)

\(=x.\left\{\left[-x-2-\left[x+3x-x-x-3\right]\right]\right\}\)

\(=x\left\{\left[-x-2-\left(4x-2x-3\right)\right]\right\}\)

\(=x\left\{\left[-x-2-\left(2x-3\right)\right]\right\}\)

\(=x\left\{-x-2-2x+3\right\}\)

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

a: Thay x=-3 vào A, ta được:

\(A=\frac{-3+2}{-3}=\frac{-1}{-3}=\frac13\)

\(x=\sqrt{\left(-3\right)^2}=\sqrt9=3\)

Thay x=3 vào A, ta được:

\(A=\frac{3+2}{3}=\frac53\)

b: \(B=\frac{3}{x+5}+\frac{20-2x}{x^2-25}\)

\(=\frac{3}{x+5}+\frac{20-2x}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{3\left(x-5\right)+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{3x-15+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{x+5}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{1}{x-5}\)

c: \(A=B\cdot\left|x-4\right|\)

=>\(\frac{x+2}{x}:\frac{1}{x-5}=\left|x-4\right|\)

=>\(\frac{\left(x+2\right)\left(x-5\right)}{x}=\left|x-4\right|\)

=>\(\begin{cases}\frac{\left(x+2\right)\left(x-5\right)}{x}\ge0\\ \left(x+2\right)^2\cdot\frac{\left(x-5\right)^2}{x^2}=\left(x-4\right)^2\end{cases}\Rightarrow\begin{cases}\left[\begin{array}{l}-2\le x<0\\ x\ge5\end{array}\right.\\ \left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\end{cases}\)

Ta có: \(\left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\)

=>\(\left(x^2-3x-10\right)^2=\left(x^2-4x\right)^2\)

=>\(\left(x^2-4x-x^2+3x+10\right)\left(x^2-4x+x^2-3x-10\right)=0\)

=>(-x+10)\(\left(2x^2-7x-10\right)=0\)

TH1: -x+10=0

=>-x=-10

=>x=10(nhận)

TH2: \(2x^2-7x-10=0\)

=>\(x^2-\frac72x-5=0\)

=>\(x^2-2\cdot x\cdot\frac74+\frac{49}{16}-\frac{129}{16}=0\)

=>\(\left(x-\frac74\right)^2=\frac{129}{16}\)

=>\(\left[\begin{array}{l}x-\frac74=\frac{\sqrt{129}}{4}\\ x-\frac74=-\frac{\sqrt{129}}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{129}+7}{4}\left(loại\right)\\ x=\frac{-\sqrt{129}+7}{4}\left(nhận\right)\end{array}\right.\)

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

\(\Leftrightarrow x-4=\sqrt{x}+2\)

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9

bạn ơi. Cho tớ hỏi là tại sao |x-4|= A/B hả bạn ?. Giải thích cho mình với

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

c,M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) :  \(\dfrac{\sqrt{x}+3}{\sqrt{x}+5}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) \(\times\) \(\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\) = \(\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}\)

 M = 1  - \(\dfrac{7}{\sqrt{x}+3}\) 

 M \(\in\) Z ⇔ 7 ⋮ \(\sqrt{x}\) + 3 vì \(\sqrt{x}\) ≥ 0 ⇒ \(\sqrt{x}\) + 3 ≥ 3 ⇒ 0< \(\dfrac{7}{\sqrt{x}+3}\) ≤ \(\dfrac{7}{3}\)

⇒ M Đạt giá trị nguyên lớn nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) đạt giá trị nguyên nhỏ nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) = 1 ⇔ \(\sqrt{x}\) + 3  = 7 ⇔ \(\sqrt{x}\) = 4 ⇔ \(x\) = 16 

Mnguyên(max)  = 1 - 1 = 0 xảy ra khi \(x\) = 16