Cho biểu thức \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x}{x-2\sqrt{x}+1}\) ( \(x>0;x\ne1\) )
1. Rút gọn biểu thức
2. Tìm giá trị của \(x\) để \(P>\dfrac{1}{2}\)
Giúp câu 2 với ạ
K
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Những câu hỏi liên quan
14 tháng 9 2021
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
14 tháng 9 2021
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
6 tháng 7 2021
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
12 tháng 10 2021
\(M=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\)
12 tháng 10 2021
\(M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
\(B=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)=\left(\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\dfrac{x+1}{\sqrt{x}}\)
Để \(B< 0\Rightarrow\dfrac{x+1}{\sqrt{x}}< 0\)
\(\Rightarrow x+1< 0\) (vô lý do \(x>0\))
Vậy ko tồn tại x thỏa mãn yêu cầu
MH
26 tháng 4 2022
Điều kiện: \(x\ge0,x\ne1\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\Rightarrow A>0\)
Lại có: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}+1>0;x+\sqrt{x}>0\) với mọi \(x\in TXĐ\)
\(\Rightarrow A-2< 0\Rightarrow A< 2\)
Vậy \(0< A< 2\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
7 tháng 7 2021
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)
7 tháng 12 2021
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
TP
8 tháng 12 2021
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
Lời giải:
1. \(P=\left[\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right]:\frac{x}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{x}=\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{x\sqrt{x}}=\frac{x-1}{x\sqrt{x}}\)
2.
\(P>\frac{1}{2}\Leftrightarrow \frac{x-1}{x\sqrt{x}}> \frac{1}{2}\)
\(\Leftrightarrow \frac{2x-2-x\sqrt{x}}{2x\sqrt{x}}>0\)
\(\Leftrightarrow 2x-2-x\sqrt{x}>0\)
\(\Leftrightarrow x\sqrt{x}+2< 2x\)
Điều này vô lý do theo BĐT Cô-si thì:\(x\sqrt{x}+2=\frac{x\sqrt{x}}{2}+\frac{x\sqrt{x}}{2}+2\geq 3\sqrt[3]{\frac{x^3}{2}}>\frac{3x}{\sqrt[3]{2}}> 2x\)
Vậy không tồn tại $x$ thỏa mãn.
1) Ta có: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{x}\)
\(=\dfrac{x-1}{x\sqrt{x}}\)