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a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(=\left[\frac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\left(\sqrt{x}+1\right)\)
\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}+1\right)=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) Ta có: \(x=\frac{4}{9}\)thỏa mãn ĐKXĐ
\(\Rightarrow\)Thay \(x=\frac{4}{9}\)vào biểu thức A ta có:
\(A=\frac{\sqrt{\frac{4}{9}}+2}{\sqrt{\frac{4}{9}}-1}=\frac{\frac{2}{3}+2}{\frac{2}{3}-1}=\frac{\frac{8}{3}}{-\frac{1}{3}}=-8\)
c) Ta có: \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow4\left(\sqrt{x}+2\right)=5\left(\sqrt{x}-1\right)\)\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)
\(\Leftrightarrow\sqrt{x}=13\)\(\Leftrightarrow x=169\)( thỏa mãn ĐKXĐ )
Vậy \(x=169\)
\(a,ĐKXĐ:x\ne1,x>0\)
\(A=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(A=\frac{3+\sqrt{x}-1}{x-1}.\frac{\sqrt{x}+1}{1}\)
\(A=\frac{2+\sqrt{x}}{\sqrt{x}-1}\)
với \(x=\frac{4}{9}\)
\(< =>A=\frac{2+\sqrt{\frac{4}{9}}}{\sqrt{\frac{4}{9}}-1}\)
\(A=\frac{2+\frac{2}{3}}{\frac{2}{3}-1}=\frac{\frac{8}{3}}{\frac{-1}{3}}=-8\)
\(c,\frac{5}{4}=\frac{2+\sqrt{x}}{\sqrt{x}-1}\)
\(5\sqrt{x}-5=8+4\sqrt{x}\)
\(\sqrt{x}=13< =>x=169\)
a, \(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}=\frac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)ĐK : \(x\ne1;x\ge0\)
\(=\frac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b, Thay \(x=\frac{9}{4}\Rightarrow\sqrt{x}=\frac{3}{2}\)vào biểu thức A ta được
\(\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{\frac{3}{2}}{\frac{1}{2}}=3\)Vậy với x = 9/4 thì A = 3
c, Ta có : \(A=\frac{9}{4}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{9}{4}\Rightarrow4\sqrt{x}=9\sqrt{x}-9\)
\(\Leftrightarrow5\sqrt{x}=9\Leftrightarrow\sqrt{x}=\frac{9}{5}\Leftrightarrow x=\frac{81}{25}\)
Vậy với A = 9/4 thì x = 81/25
\(ĐKXĐ=x\ne1;x>0\)
\(A=\frac{\sqrt{x}^3+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(A=\frac{\sqrt{x}^3+1-\sqrt{x}^3+\sqrt{x}+x-1}{x-1}\)
\(A=\frac{\sqrt{x}+x}{x-1}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(b,A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{3}{\frac{2}{\frac{1}{2}}}=3\)
\(c,\frac{5}{4}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(5\sqrt{x}-5=4\sqrt{x}\)
\(\sqrt{x}=5< =>x=25\)
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
\(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(A=\frac{x\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(x-\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)}\)
b)Khi \(x=\frac{9}{4}\)
\(\Rightarrow\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=3\)
c)\(A=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)}< 1\)
\(\Leftrightarrow\sqrt{x}< \sqrt{x}-1\)(Voly)
=>ko có giá trị nào
\(a,ĐKXĐ:x\ge0;x\ne1\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(P=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(P=\left(x+1\right)^2\left(x-1\right)^2\)
\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)
\(P=\left(x^2+x-x-1\right)^2\)
\(P=\left(x^2-1\right)^2\)
b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)
\(\left(2-\sqrt{3}\right)^2\)
\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)
\(x^2-1< 2-\sqrt{3}\)
\(x^2< 3-\sqrt{3}\)
\(x< \sqrt{3-\sqrt{3}}\)
a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có: \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)}-\sqrt{x}\right)\)
\(P=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2=\left(x-1\right)^2\)
b) Với x > = 0 và x khác 1
Ta có: \(P< 7-4\sqrt{3}\)
<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)
<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)
<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)
<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)
<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)
Trả lời
a)ĐKXĐ
x > = 0 ; x khác 4
P=\(\frac{\sqrt{x}-\sqrt{x}-2}{\left(\sqrt{x-2}\right)\left(\sqrt{x-2}\right)}=\)\(\frac{-2}{x-4}\)
b)P=1/5
=>\(\frac{-2}{x-4}=\frac{1}{5}\Rightarrow-10=x-4\Rightarrow x=-6\)(loại vì x > 0)
Vậy không có x
a, \(\hept{\begin{cases}x-1\ne0\\\sqrt{x}+1\ge1\end{cases}\Rightarrow\hept{\begin{cases}x\ne1\\\sqrt{x}\ge0\end{cases}\Rightarrow}x>1}\)
=> ĐKXĐ: x>1