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a: \(S=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-4-x+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: Để S=0 thì \(\sqrt{x}-2=0\)
hay x=4(loại)
a) điều kiện xác định : \(x\ge0;x\ne\dfrac{9}{4}\)
ta có : \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)b) thay \(x=\dfrac{3-2\sqrt{2}}{4}=\left(\dfrac{\sqrt{2}-1}{2}\right)^2\) vào \(P\) ta có :
\(P=\dfrac{3\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}-5}{2\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}+1}=\dfrac{3\left(\dfrac{\sqrt{2}-1}{2}\right)-5}{2\left(\dfrac{\sqrt{2}-1}{2}\right)+1}=\dfrac{6-13\sqrt{2}}{4}\)
c) ta có : \(P-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5-3\sqrt{x}-\dfrac{3}{2}}{2\sqrt{x}+1}\)
\(=\dfrac{\dfrac{-13}{2}}{2\sqrt{x}+1}< 0\) \(\Rightarrow P< \dfrac{3}{2}\)
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}+\dfrac{\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\dfrac{\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Khi x=17-12 căn 2 thì \(A=\dfrac{17-12\sqrt{2}+3-2\sqrt{2}+1}{3-2\sqrt{2}}=7\)
a: \(A=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b: \(A-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}-\dfrac{3}{2}\)
\(=\dfrac{6\sqrt{x}-10-6\sqrt{x}-3}{2\left(2\sqrt{x}+1\right)}=\dfrac{-13}{2\left(2\sqrt{x}+1\right)}< 0\)
=>A<3/2
a: \(S=\dfrac{x+1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(x+1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{\left(4-2\sqrt{3}+1\right)\left(\sqrt{3}-1+1\right)}{\left(\sqrt{3}-1\right)\left(4-2\sqrt{3}-1\right)}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\cdot\sqrt{3}}{\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}\)
a) điều kiện \(x>0;x\ne1\)
\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}+1\right)^2+\left(1-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1+\sqrt{x}-1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}=\dfrac{x+1}{4\sqrt{x}}\)
Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
a: \(B=\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)
\(=\left(2x+\sqrt{x}-1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=-\dfrac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(A=\dfrac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\dfrac{-x+\sqrt{x}-1}{\sqrt{x}}\)
b: Khi \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) thì
\(A=\dfrac{-17+12\sqrt{2}+3-2\sqrt{2}-1}{3-2\sqrt{2}}=-5\)
c: \(A=\dfrac{-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{\sqrt{x}}< 0\)
=>căn A không tồn tại
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\)
a )
\(S=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b )
\(S=3\)
\(\Leftrightarrow x-\sqrt{x}+1=3\)
\(\Leftrightarrow x-\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=4\end{matrix}\right.\)
Vậy \(x=4\)