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\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)(ĐK: \(x\ge0,x\ne1\))
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\frac{5}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{5}{\sqrt{x}}\)
\(\Rightarrow x=5\left(x+\sqrt{x}+1\right)\)
\(\Leftrightarrow4x+5\sqrt{x}+1=0\)(vô nghiệm do \(x\ge0\))
\(A-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)(vì \(x\ne1\))
Do đó \(A< \frac{1}{3}\).
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}}-\frac{1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
Đặt \(a+b=x\) , \(ab=y\)
Ta có biểu thức cần rút gọn :
\(\frac{1}{x^3}.\frac{x\left(x^2-3y\right)}{y^3}+\frac{3}{x^4}.\frac{x^2-2y}{y^2}+\frac{6}{x^5}.\frac{x}{y}=\frac{x^4-3x^2y+3yx^2-6y^2+6y^2}{x^4y^3}=\frac{x^4}{x^4y^3}=\frac{1}{y^3}=\frac{1}{a^3b^3}\)
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a-1}}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà \(\sqrt{a}-1< \sqrt{a}\) => \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy M < 1.
Nếu đề đúng:
Sử dụng liên hợp để trục căn thức ở mẫu:
\(\frac{1}{\sqrt{1}+\sqrt{5}}=\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{\sqrt{5}-1}{5-1}=\frac{\sqrt{5}-1}{4}\)
Tương tự như vậy ta sẽ có:
\(N=\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\sqrt{13}-\sqrt{9}}{\left(\sqrt{13}-\sqrt{9}\right)\left(\sqrt{13}+\sqrt{9}\right)}+\frac{\sqrt{17}-\sqrt{13}}{\left(\sqrt{17}-\sqrt{13}\right)\left(\sqrt{17}+\sqrt{13}\right)}\)
\(+\frac{\sqrt{21}-\sqrt{17}}{\left(\sqrt{21}-\sqrt{17}\right)\left(\sqrt{21}+\sqrt{17}\right)}+\frac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)
\(=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+\frac{\sqrt{17}-\sqrt{13}}{4}+\frac{\sqrt{21}-\sqrt{17}}{4}+\frac{\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1+\sqrt{13}-\sqrt{9}+\sqrt{17}-\sqrt{13}+\sqrt{21}-\sqrt{17}+\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1-\sqrt{9}+\sqrt{21}+\sqrt{25}-\sqrt{23}}{4}=\frac{\sqrt{5}-1-3+\sqrt{21}+5-\sqrt{23}}{4}=\frac{1+\sqrt{5}+\sqrt{21}-\sqrt{23}}{4}\)
a/
\(1995^n.1997^n=\left(1995.1997\right)^n\)
\(1996^{2n}=\left(1996^2\right)^n\)
\(1995.1997=\left(1996-1\right).\left(1996+1\right)=1996^2-1\)
\(\Rightarrow1995.1997< 1996^2\Rightarrow1995^n.1997^n< 1996^{2n}\)
b/
\(A=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{9.20}+\frac{1}{9.30}+\frac{1}{9.42}+\frac{1}{9.56}\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(A=\frac{1}{9}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{9}\right)=\frac{1}{9}.\frac{8}{9}=\frac{8}{81}\)