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1) ĐK: \(x\ge0;x\ne1\)
2) \(E=\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}=\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
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 \(\sqrt{x-1}-1\ge-1>-4\))
\(\Leftrightarrow x=26\)
\(A=\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(ĐKXĐ:x\ge0;x\ne1\right)\)
\(< =>A=\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1}{x-\sqrt{x}}+\sqrt{x}\)
\(< =>A=\frac{1+\sqrt{x}\left(x-\sqrt{x}\right)}{x-\sqrt{x}}=\frac{1+x\sqrt{x}-x}{x-\sqrt{x}}\)
Với \(x=\frac{18}{4+\sqrt{7}}\)thì \(A=\frac{1+\frac{18}{4+\sqrt{7}}.\sqrt{\frac{18}{4+\sqrt{7}}}-\frac{18}{4+\sqrt{7}}}{\frac{18}{4+\sqrt{7}}-\sqrt{\frac{18}{4+\sqrt{7}}}}\)
\(=\frac{1}{18+\frac{4}{7}-\sqrt{18+\frac{4}{7}}}+\sqrt{18+4\sqrt{7}}\)
Em mới lớp 7 nên chỉ làm được thế thôi ạ :3
\(E=\frac{1}{\sqrt{x+2\sqrt{x-1}}}+\frac{1}{\sqrt{x-2\sqrt{x-1}}}\)
\(=\frac{1}{\sqrt{x-1+2\sqrt{x-1}+1}}+\frac{1}{\sqrt{x-1-2\sqrt{x-1}+1}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}+\frac{1}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)
\(=\frac{1}{\left|\sqrt{x-1}+1\right|}+\frac{1}{\left|\sqrt{x-1}-1\right|}\)
Quy dong tinh tiep ! ok