\(\left(\dfrac{2x-1}{\sqrt{x^3}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)rút gọn giúp mik bài này nha mn
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KV
3 tháng 6 2023
a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)
\(=3\sqrt{x}-6\)
b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\) (1)
ĐKXĐ: \(x>0\)
\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)
\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)
\(\Leftrightarrow3x-10\sqrt{x}+1=0\) (2)
Đặt \(t=\sqrt{x}\ge0\)
\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)
\(\Delta'=25-4=22\)
Phương trình có hai nghiệm phân biệt:
\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)
\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)
Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)
Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)
Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
2 tháng 6 2023
a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)
b: P=(4căn x-1)/căn x
=>3x-6căn x-4căn x+1=0
=>3x-10căn x+1=0
=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9
24 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+1\right)\)
\(:\left(1-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+\dfrac{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(:\left(\dfrac{\text{}\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(.\left(\dfrac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(A=1\)
15 tháng 5 2021
\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((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`
23 tháng 8 2021
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
23 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
30 tháng 11 2021
\(=\dfrac{2x-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x\left(x-1\right)}{x+\sqrt{x}+1}\)
Bạn @Phùng Khánh Linh làm hơi tắt rồi để mình giúp bạn làm cho dễ hiểu nhá !
\(\left(\dfrac{2x-1}{\sqrt{x^3}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-3}{x+\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\left(\dfrac{2x-1}{\sqrt{x^3}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x+4}{x+\sqrt{x}+1}\right)=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-3}{x+\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)