Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
Điều kiện xác định : \(0\le x\ne1\)
- \(H=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(x-1\right)-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=-2\sqrt{x-1}+x=\left(x-1-2\sqrt{x-1}+1\right)=\left(\sqrt{x-1}-1\right)^2\)
- Với \(x=\frac{53}{9-2\sqrt{3}}\) tính H kết quả rất lẻ.
- H = 16 \(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=16\Leftrightarrow\left|\sqrt{x-1}-1\right|=4\)
\(\Leftrightarrow\sqrt{x-1}-1=4\) (Vì \(\sqrt{x-1}-1\ge-1>-4\))
\(\Leftrightarrow x=26\)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(H=\frac{4}{\sqrt{x}+1}+\frac{2}{1-\sqrt{x}}-\frac{\sqrt{x}-5}{x-1}\)
\(=\frac{4}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(4\sqrt{x}-4\right)-\left(2\sqrt{x}+2\right)-\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)
\(H=\frac{4}{\sqrt{x}+1}+\frac{2}{1-\sqrt{x}}-\frac{\sqrt{x}-5}{x-1}\)
\(=\frac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)