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
\(ĐKXĐ=x\ne1;x>0\)
\(A=\frac{\sqrt{x}^3+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(A=\frac{\sqrt{x}^3+1-\sqrt{x}^3+\sqrt{x}+x-1}{x-1}\)
\(A=\frac{\sqrt{x}+x}{x-1}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(b,A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{3}{\frac{2}{\frac{1}{2}}}=3\)
\(c,\frac{5}{4}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(5\sqrt{x}-5=4\sqrt{x}\)
\(\sqrt{x}=5< =>x=25\)
nhầm, nhìn phần c thành b nhé
c, Ta có :
\(A=\frac{5}{4}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}=5\sqrt{x}-5\)
\(\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\)
Vậy với A = 5/4 thì x = 25