Rút gọn các biểu thức

Giải SBT Toán 9 Bài 4: Liên hệ giữ phép chia và phép khai phương

Bài 41 trang 11 Sách bài tập Toán 9 Tập 1: Rút gọn các biểu thức:

a) $\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge 0\right)$

b)$\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{y-2\sqrt{y}+1}{{\left(x-1\right)}^4}}\left(x\ne 1,y\ne 1,y\ge 0\right)$

Lời giải:

a) $\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{{\left(\sqrt{x}\right)}^2-2.\sqrt{x}.1+1^2}{{\left(\sqrt{x}\right)}^2+2.\sqrt{x}.1+1^2}}$

 $=\sqrt{\frac{{\left(\sqrt{x}-1\right)}^2}{{\left(\sqrt{x}+1\right)}^2}}=\frac{\sqrt{{\left(\sqrt{x}-1\right)}^2}}{\sqrt{{\left(\sqrt{x}+1\right)}^2}}=\frac{\left.\sqrt{x}-1\right|}{\left.\sqrt{x}+1\right|}$

  = $\frac{\left.\sqrt{x}-1\right|}{\sqrt{x}+1}$(Vì $\sqrt{x}+1>0$ với mọi x)

+ Nếu $0\le x<1$ thì

$\sqrt{x}-1<0\Rightarrow \left.\sqrt{x}-1\right|=1-\sqrt{x}$

Do đó:$\frac{\left.\sqrt{x}-1\right|}{\sqrt{x}+1}$= $\frac{1-\sqrt{x}}{\sqrt{x}+1}$

+ Nếu $x\ge 1$ thì $\sqrt{x}-1\ge 0$

$\Rightarrow \left.\sqrt{x}-1\right|=\sqrt{x}-1$

Do đó: $\frac{\left.\sqrt{x}-1\right|}{\sqrt{x}+1}$ = $\frac{\sqrt{x}-1}{\sqrt{x}+1}$

b) $\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{y-2\sqrt{y}+1}{{\left(x-1\right)}^4}}$

$=\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{{\left(\sqrt{y}\right)}^2-2.\sqrt{y}.1+1}{{\left(x-1\right)}^4}}=\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{{\left(\sqrt{y}-1\right)}^2}{{\left(x-1\right)}^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{\sqrt{{\left(\sqrt{y}-1\right)}^2}}{\sqrt{{\left(x-1\right)}^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{\left.\sqrt{y}-1\right|}{{\left(x-1\right)}^2}=\frac{\left.\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}$

+ Nếu $0<y<1$thì 

$\sqrt{y}-1<0\Rightarrow \left.\sqrt{y}-1\right|=-\left(\sqrt{y}-1\right)$

Do đó:

$\frac{\left.\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{-\left(\sqrt{y}-1\right)}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{-1}{x-1}$

+ Nếu y > 1 thì

$\sqrt{y}-1>0\Rightarrow \left.\sqrt{y}-1\right|=\sqrt{y}-1$

Do đó:

$\frac{\left.\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{\left(\sqrt{y}-1\right)}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{1}{x-1}$