\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
Giups em giải nhanh bài này với ạ! Em cảm ownnhieeuf ạ
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9 tháng 3 2022
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
16 tháng 2 2022
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
24 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+1\right)\)
\(:\left(1-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+\dfrac{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(:\left(\dfrac{\text{}\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(.\left(\dfrac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(A=1\)
5 tháng 8 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-1}{\sqrt{x}+1}\)
Thay x=3 vào B, ta được:
\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)
HD
26 tháng 7 2021
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
10 tháng 11 2021
\(ĐK:x\ge0;x\ne1;x\ne4\\ P=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\\ P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+4}\\ P=\dfrac{2\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-4}{3\sqrt{x}+3}\)
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`
\(B=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x+2}{\sqrt{x}}\)
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