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7 tháng 1 2023
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\cdot\dfrac{x-4}{4}=\dfrac{2\sqrt{x}}{4}=\dfrac{1}{2}\sqrt{x}\)
2 tháng 10 2021
\(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\left(x>0;x\ne1;x\ne\dfrac{1}{4}\right)\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
2 tháng 10 2021
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
D
A= (2/x-√x - 1/√x-1) : x-4/x√x+√x - 2x với x>0, x khác 1, x khác 4 a) rút gọn A b) tìm x để A > -1/2
1
5 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{2}{x-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-4}{x\sqrt{x}+\sqrt{x}-2x}\)
\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{x-4}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\sqrt{x}+1}{\sqrt{x}+2}\)
4 tháng 10 2021
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
3 tháng 10 2021
Câu a đã làm: F=(2√x/2√x-1 - 1/√x) ( √x+1/√x-1 + 3x/x-2√x+1) với x >0, x khác 1, x khác 1/4 a) rút gọn F - Hoc24
\(b,F=2\Leftrightarrow\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow2\sqrt{x}\left(x-2\sqrt{x}+1\right)=2x\sqrt{x}-4x+2\sqrt{x}+2x-2\sqrt{x}+1\\ \Leftrightarrow2x\sqrt{x}-4x+2\sqrt{x}=2x\sqrt{x}-2x+1\\ \Leftrightarrow2x-2\sqrt{x}+1=0\\ \Leftrightarrow2\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\\ \Leftrightarrow x\in\varnothing\)
24 tháng 8 2023
1: A=2
=>\(\sqrt{x}+1=2\left(\sqrt{x}-2\right)\)
=>\(2\sqrt{x}-4=\sqrt{x}+1\)
=>\(\sqrt{x}=5\)
=>x=25
2: A<1
=>A-1<0
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>0<=x<4
3: A<1/3
=>A-1/3<0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{1}{3}< 0\)
=>\(\dfrac{3\sqrt{x}+3-\sqrt{x}+2}{3\left(\sqrt{x}-2\right)}< 0\)
=>\(\dfrac{2\sqrt{x}+5}{3\left(\sqrt{x}-2\right)}< 0\)
=>\(\sqrt{x}-2< 0\)
=>0<=x<4
4:
A=căn x
=>\(\sqrt{x}+1=x-2\sqrt{x}\)
=>\(x-3\sqrt{x}-1=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{13}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{3-\sqrt{13}}{2}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{11+3\sqrt{13}}{2}\)
ĐKXĐ: x>0; x<>4
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{2-\sqrt{x}}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)