a:Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

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

b: Ta có: \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(x-3\sqrt{x}+2\right)\)

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

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

???

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1))))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

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

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

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

\(Q=\left(1-\dfrac{1}{\sqrt{x}}\right)^2:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: x > 0)

\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)^2:\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)

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

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

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

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

\(Q=\dfrac{x-1}{x}\)

\(Q=\left(1-\dfrac{1}{\sqrt{x}}\right)^2:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)(ĐKXĐ: x > 0)

\(=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)^2:\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x}:\left[\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)

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

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

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

\(Q=\left(x-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\cdot\dfrac{x+3\sqrt{x}+1}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ Q=\left(x-x-\sqrt{x}-1\right)\cdot\dfrac{x+3\sqrt{x}+1}{\sqrt{x}+1}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(x+3\sqrt{x}+1\right)}{\sqrt{x}+1}=-x-3\sqrt{x}-1\)

a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))

Vì hai vế ko âm, bp 2 vế ta được:

2x² - 3 = 4x - 3

\(\Leftrightarrow\) 2x² = 4x

\(\Leftrightarrow\) x² = 2x

\(\Leftrightarrow\) x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy S = {2}

b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)

Vì hai vế ko âm, bp 2 vế ta được:

2x - 1 = x - 1

\(\Leftrightarrow\) x = 0 (KTM)

Vậy x = \(\varnothing\)

c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)

Vì hai vế ko âm, bp 2 vế ta được:

x² - x - 6 = x - 3

\(\Leftrightarrow\) x² - 2x - 3 = 0

\(\Leftrightarrow\) x² - 3x + x - 3 = 0

\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy S = {3}

d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))

Vì hai vế ko âm, bp 2 vế ta được:

x² - x = 3x - 5

\(\Leftrightarrow\) x² - 4x + 5 = 0

\(\Leftrightarrow\) x² - 4x + 4 + 1 = 0

\(\Leftrightarrow\) (x - 2)² + 1 = 0

Vì (x - 2)² \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)² + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!

Nguyễn Lê Phước Thịnh nhờ anh xíu ạ

\(\sqrt{\left(120-11\right)^2}+\sqrt{\left(10-\sqrt{120}\right)^2}\)

\(=120-11+10+\sqrt{120}\)

\(=\sqrt{120}\left(\sqrt{120}+1\right)-1\)

\(a,=\left(120-11\right)+\left|10-\sqrt{120}\right|=109+\sqrt{120}-10=99+2\sqrt{30}\\ b,=\sqrt{\left(\sqrt{x+1}+1\right)^2-\left(\sqrt{x+1}+1\right)^2}=\sqrt{0}=0\)

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