2. rút gọn biểu thức
P=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
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Những câu hỏi liên quan
3 tháng 3 2021
Ta có: \(P=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\cdot\dfrac{1}{1-\sqrt{x}}\)
\(=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\cdot\dfrac{1}{1-\sqrt{x}}\)
\(=\sqrt{x}+1\)
DH
3 tháng 3 2021
\(P=\left(\dfrac{x}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right):\left(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\right)\)
\(P=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x-1\right)}+\dfrac{\left(\sqrt{x}-1\right)}{\left(x-1\right)}+\dfrac{2}{x-1}\right):\left(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\right)\)
\(P=\dfrac{x+\sqrt{x}+\sqrt{x}-1+2}{x-1}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=\dfrac{x+2\sqrt{x}+1}{x-1}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(P=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
DH
2 tháng 3 2021
\(P=\left(\dfrac{2\sqrt{x}+2}{x\sqrt{x}-\sqrt{x}+x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{1}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2}{\left(x-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2}{\left(x-1\right)}-\dfrac{\left(\sqrt{x}+1\right)}{\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{1-\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)=-\left(\sqrt{x}+2\right)\)
NM
2 tháng 3 2021
ĐKXĐ x\(\ge0;x\ne1\)
P=\(\left(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{1}{\sqrt{x}+1}\right)\)
P=\(\left(\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
P=\(\dfrac{2-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
P=\(\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{-1}{\sqrt{x}+2}\)
3 tháng 3 2021
Ta có: \(P=\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{x\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)\(=\dfrac{3\sqrt{x}-6+\sqrt{x}+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)^2}\)
DH
2 tháng 3 2021
\(P=\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{-6}{x-4}\right):\left(\dfrac{6}{\sqrt{x}+2}\right)=\dfrac{-1}{\sqrt{x}-2}\)
NM
2 tháng 3 2021
ĐKXĐ x\(\ge0,x\ne1,x\ne4\)
P=
P=\(\left(\dfrac{\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}+\dfrac{x+2\sqrt{x}+4}{x-1}\right):\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+\sqrt{x}+1+2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
P=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
P=\(\dfrac{x-1+\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
P=\(\dfrac{x\sqrt{x}+x-9}{3\left(x-3\right)}\)
3 tháng 3 2021
Ta có: \(P=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
\(=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\dfrac{-x+9+2x-4\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
GH
3 tháng 7 2023
a
ĐK: \(1< x\ne10\)
Đặt \(t=\sqrt{x-1}\Rightarrow x=t^2+1;0< t\ne3\)
Khi đó:
\(P=\left(\dfrac{t}{3+t}+\dfrac{t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t^2-3t}-\dfrac{1}{t}\right)\\ =\left(\dfrac{t\left(3-t\right)+t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t\left(t-3\right)}-\dfrac{1}{t}\right)\\ =\dfrac{3t+9}{\left(3-t\right)\left(3+t\right)}:\dfrac{3t+1-t+3}{t\left(t-3\right)}=\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}:\dfrac{2t+4}{t\left(t-3\right)}\\ =\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}.\dfrac{t\left(t-3\right)}{2t+4}=\dfrac{-3t}{2t+4}=\dfrac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
b
Ta có:
\(x=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{5}\left|1-\sqrt{2}\right|\)
\(=\sqrt{2}+1-\left(\sqrt{5}+1\right)\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\)
\(=\sqrt{2}+1-\sqrt{5}\left|1-\sqrt{2}\right|-\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\\ =\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)
Vậy \(P=\dfrac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\dfrac{1}{2}\)
3 tháng 3 2021
Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+3}{x-2\sqrt{x}-3}-\dfrac{x}{x-3\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{3-\sqrt{x}}\)
\(=\left(\dfrac{x\sqrt{x}-2x+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
\(=\dfrac{x\sqrt{x}-2x+3\sqrt{x}-x\sqrt{x}-x}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3x+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3}{\sqrt{x}+1}\)
P = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)DKXD: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
= \(\sqrt{x}+\sqrt{x}\)
= \(2\sqrt{x}\)
Vậy tại x ∈ ĐKXĐ thì P = \(2\sqrt{x}\)
Ta có: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}+\sqrt{x}\)
\(=2\sqrt{x}\)