Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

ĐKXĐ: \(x\ge0;x\ne4\)

\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot...\cdot\dfrac{400}{399}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{20\cdot20}{19\cdot21}\)
\(=\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot20\cdot20}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot20}{1\cdot2\cdot3\cdot...\cdot19}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot20}{3\cdot4\cdot5\cdot...\cdot21}\)
\(=20\cdot\dfrac{2}{21}\)
\(=\dfrac{40}{21}\)

bạn đợi mình xíu, mình đang trình bày ^^

Bài giải đây nha bạn https://imgur.com/gallery/NAS59mp

ĐKXĐ: \(x\ge\frac{1}{2}\)

Chắc pt là thế này:

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=3\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=3\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=3\)

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=3\)

- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\)

\(\Leftrightarrow\sqrt{x-1}+1+\sqrt{x-1}-1=3\)

\(\Leftrightarrow\sqrt{x-1}=\frac{3}{2}\Rightarrow x=\frac{13}{4}\) (t/m)

- Nếu \(\frac{1}{2}\le x< 2\)

\(\Leftrightarrow\sqrt{x-1}+1+1-\sqrt{x-1}=3\Leftrightarrow2=3\) (vô lý)

Vậy pt có nghiệm duy nhất \(x=\frac{13}{4}\)

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

=3x^4+9x^3+3x^2

\(\left\{{}\begin{matrix}y-\dfrac{2}{5}=\dfrac{x}{50}\\y+1=\dfrac{x}{40}\end{matrix}\right.\)

`=> y -2/5 -y-1 = x/50 -x/40`

`<=> -7/5 = x(1/50-1/40)`

`=> x= -7/5 : (1/50 -1/40) `

`<=> x =280`

`=> y +1 =280/40 = 7`

`<=> y = 6`

Vậy.....