Bài 8: Rút gọn biểu thức chứa căn bậc hai

Lời giải:

ĐKXĐ: $a\geq 0; a\neq 4$

\(A=\left[\frac{\sqrt{a}(\sqrt{a}-2)-\sqrt{a}(\sqrt{a}+2)}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}\right].(\sqrt{a}+2)\)

\(=\frac{-4\sqrt{a}+4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{1}{2-\sqrt{a}}\)

`A=((x\sqrtx-1)/( x-\sqrtx)-(x\sqrtx+1)/(x+\sqrtx)(\sqrtx/(\sqrtx-1)-1/(\sqrtx+1))(x>0,x ne 1)`
`=(((\sqrtx-1)(x+\sqrtx+1))/( x-\sqrtx)-((\sqrtx+1)(x-\sqrtx+1))/(x+\sqrtx)((x+\sqrtx-\sqrtx+1)/(x-1))`
`=((x+\sqrtx+1+x-\sqrtx+1)/\sqrtx) .((x+1) /( x-1)) `
`=((2x+2)/\sqrtx).((x+1) /(x-1 ) )`
`=( 2(x+1)^2) /(\sqrtx(x-1))`

`A=((x\sqrtx-1)/( x-\sqrtx)-(x\sqrtx+1)/(x+\sqrtx)(\sqrtx/(\sqrtx-1)-1/(\sqrtx+1))(x>0,x ne 1)`
`=(((\sqrtx-1)(x+\sqrtx+1))/( x-\sqrtx)-((\sqrtx+1)(x-\sqrtx+1))/(x+\sqrtx))((x+\sqrtx-\sqrtx+1)/(x-1))`
`=((x+\sqrtx+1+x-\sqrtx+1)/\sqrtx) .((x+1) /( x-1)) `
`=((2x+2)/\sqrtx).((x+1) /(x-1 ) )`
`=( 2(x+1)^2) /(\sqrtx(x-1))`

\(\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\) Đk: \(\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)

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

= \(\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

= \(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

\(\left(\dfrac{x-4}{\sqrt{x}+2}+\dfrac{9}{\sqrt{x}+4}\right).\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-1\right)(x \geq 0\)
`=(\sqrtx-2+9/(\sqrtx+4)).(-1/(\sqrtx+1))`
`=(x+2\sqrtx+1)/(\sqrtx+4).(-1/(\sqrtx+1))`
`=(-\sqrtx-1)/(\sqrtx+4)`

Hmmm bài này nhìn như này lẻ với khó đối với hs đại trà 9 nên mình nghĩ là `A=(x\sqrtx+1)/(x-1)-(x-1)/(\sqrtx-1)` hợp lý hơn.

`đkxđ;x>=0,x ne 1`

`A=((\sqrtx+1)(x-\sqrtx+1))/(x-1)-(x-1)/(\sqrtx-1)`

`=(x-\sqrtx-1)/(\sqrt-1)-(x-1)/(\sqrtx-1)`

`=(x-\sqrtx-1-x+1)/(\sqrtx-1)`

`=(-\sqrtx)/(\sqrtx-1)`

Ta có: \(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

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

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

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

\(C=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}-\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

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

\(=\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

Ta có: \(A=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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