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Ta thấy: k thuộc N* nên \(\sqrt{k+1}>\sqrt{k}\)
\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{2}{\left(2\sqrt{k+1}\right).\left(\sqrt{k+1}.\sqrt{k}\right)}< \frac{2}{\left(\sqrt{k+1}.\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}\)
\(=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}.\sqrt{k}\right)\left(k+1-k\right)}=2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)(đpcm).
\(\frac{1}{\sqrt{k}\left(k+1\right)}=\frac{1}{\sqrt{k+1}}.\frac{1}{\sqrt{k}\sqrt{k+1}}=\frac{1}{\sqrt{k+1}}.\frac{k+1-k}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k+1}}\left(\frac{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k}\sqrt{k+1}}\right)\)
\(=\frac{\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k}\sqrt{k+1}}.\frac{\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}}<\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}.2\)
Đề đúng sory nhé
ĐKXĐ: a > 0
a/ \(K=\left[\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{1}{\sqrt{a}-1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\left[\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+3}\right]\) \(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
b/ Ta có: \(\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
\(K=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}=\frac{\left(\sqrt{2}+2\right)\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+4\right)}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}\left(\sqrt{2}+1\right)\left(2\sqrt{2}+1\right)}\)
\(=\frac{\sqrt{2}}{2\sqrt{2}+1}\)
c/ \(K< 0\Leftrightarrow\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)\(\Rightarrow\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)< 0\)
\(\Rightarrow\sqrt{a}-1< 0\) (vì \(\left(\sqrt{a}+1\right)^2>0\)) \(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)
Vậy \(0< a< 1\) thì K < 0
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c = 3
Chứng minh rằng với mọi k > 0 ta luôn có....
.
Cho a,b,c là các số thực dương thỏa mãn a+b+c = 3
Chứng minh rằng với mọi k > 0 ta luôn có
ta có \(\left(1+\frac{1}{k}-\frac{1}{k-1}\right)^2\)
= \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)\(+\frac{2}{k-1}-\frac{2}{k}-\frac{2}{k\left(k-1\right)}\)
=\(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2k-2k+2-2}{k\left(k-1\right)}\)
= \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
=> \(\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}\)= \(1+\frac{1}{k-1}-\frac{1}{k}\)(đpcm)
CÂU CỦA BẠN KIA SAI R
bạn ấy bị sai cái phần mà cộng cho cả tử và mẫu cho a/k