Bài 1:
\(C=\left(\frac{1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\) với x>0, x khác 1
a, Rút gọn C
b, Tìm x để \(D=\sqrt{x}-C\) nguyên
Bài 2:
\(E=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(F=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right)\):\(\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
a, Tính E khi x=25
b, Rút gọn F
c, Tìm x thuộc Z để \(G=\left(-\sqrt{x}+1\right)\) F thuộc Z
Bài 1:
a) Ta có: \(C=\left(\frac{1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(=\frac{1+\sqrt{x}}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
\(=\frac{\sqrt{x}+1}{x+1}:\left(\frac{x+1-2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
\(=\frac{\sqrt{x}+1}{x+1}:\frac{x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{\sqrt{x}+1}{x+1}\cdot\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-1\)
\(=-1\)
b) Ta có: \(D=\sqrt{x}-C\)
\(=\sqrt{x}-\left(-1\right)=\sqrt{x}+1\)
Để D nguyên thì \(\sqrt{x}+1\) nguyên
mà 1 nguyên
nên \(\sqrt{x}\) nguyên
hay \(x\ge0\)
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Bài 2:
a) Thay x=25 vào biểu thức \(E=\frac{x+\sqrt{x}+1}{\sqrt{x}}\), ta được:
\(E=\frac{25+\sqrt{25}+1}{\sqrt{25}}=\frac{25+5+1}{5}=\frac{31}{5}=6,2\)
Vậy: Khi x=25 thì E=6,2
b) Ta có: \(F=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{6\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\)
\(=\left(\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{x-4+10-x}{\sqrt{x}+2}\)
\(=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{6}\)
\(=\frac{-1}{\sqrt{x}-2}\)
