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\(\dfrac{2}{xy}-\dfrac{2}{y\left(x+y\right)}-\dfrac{2}{x\left(x+y\right)}=\dfrac{2\left(x+y\right)-2x-2y}{xy\left(x+y\right)}=0\)
\(A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{\left(x+y\right)^2}}\)
\(=\sqrt{\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{y}\right)^2+\left(\dfrac{1}{x+y}\right)^2+2\times\dfrac{1}{x}\times\dfrac{1}{y}-2\times\dfrac{1}{y}\times\dfrac{1}{x+y}-2\times\dfrac{1}{x}\times\dfrac{1}{x+y}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right)}\)
\(=\left|\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right|\left(\text{đ}pcm\right)\)
ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}.\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{1+2\sqrt{a}+a}{\sqrt{a}+1}.\frac{1-2\sqrt{a}+a}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}\)
\(=\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}\right)\left(\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\frac{a-2\sqrt{a}+1}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.-\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\)
\(=1-a\)
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
Thay \(x=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\) vào chỗ \(\sqrt{x^2-1}\)
\(\Leftrightarrow2\sqrt{a\left(a+1\right)}-2a< 1\)
Lại có:\(2\sqrt{a\left(a+1\right)}\le a+a+1=2a+1\)
\(\Rightarrow2\sqrt{a\left(a+1\right)}-2a\le2a+1-2a=1\)
Dấu "=" không xảy ra
\(\Rightarrow\sqrt{a+1}-\sqrt{a}< \dfrac{1}{2\sqrt{a}}\)(đpcm)
Tại sao dấu "=" không xảy ra ?