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\(\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\frac{16}{\frac{16}{x^2}-\frac{8}{x}+1}}\)\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\left(\frac{4}{x}-1\right)^2}\)
\(\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\left(\frac{4}{x}-1\right)^2}\)\(=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\left(\frac{4-x}{x}\right)^2}\)
\(=\frac{2\sqrt{x-4}}{\left(\frac{4-x}{x}\right)^2}=\frac{2x^2\sqrt{x-4}}{\left(x-4\right)^2}=\frac{2x^2}{\sqrt{x-4}^3}\)
bài bạn YIM YIM sai nhé, mk làm lại và chỉnh lại đề luôn, bạn tham khảo:
ĐK: \(x>4\)
\(A=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\frac{16}{x^2}-\frac{8}{x}+1}\)
\(=\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\left(1-\frac{4}{x}\right)^2}\)
\(=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{\left(\frac{x-4}{x}\right)^2}\)
Nếu \(4< x\le8\)thì:
\(A=\frac{\sqrt{x-4}+2+2-\sqrt{x-4}}{\left(\frac{x-4}{x}\right)^2}\)
\(=\frac{4x^2}{\left(x-4\right)^2}\)
Nếu \(x>8\)thì:
\(A=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\frac{\left(x-4\right)^2}{x^2}}=\frac{2x^2}{\sqrt{x-4}^3}\)
Rút Gọn:
\(A=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\frac{8}{x}+\frac{16}{x^2}}}\)
\(=\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\sqrt{\left(\frac{4}{x}-1\right)^2}}\)
\(=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\frac{4}{x}-1}\)
\(=\frac{2\sqrt{x-4}}{\frac{4-x}{x}}\)
\(=-\frac{2x\sqrt{x-4}}{x-4}\)
\(=\frac{-2x}{\sqrt{x-4}}\)
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:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
Trả lời:
\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(ĐK:x\ge0;x\ne16\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)
\(A=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\sqrt{\left(\frac{4}{x}-1\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\sqrt{\left(1-\frac{4}{x}\right)^2}}=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\frac{4}{x}}\)
- Với \(x\ge8\Rightarrow\sqrt{x-4}-2\ge0\)
\(\Rightarrow A=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\frac{x-4}{x}}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}\)
- Với \(4< x\le8\)
\(\Rightarrow A=\frac{\sqrt{x-4}+2+2-\sqrt{x-4}}{\frac{x-4}{x}}=\frac{4x}{x-4}\)