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1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(\sqrt{a}+\sqrt{b}+\sqrt{c}=3< =>\left(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\right)=9< =>\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=2\\
\\
\)
Ở đâu có 2 thì thay vào @@
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\left(a+b+c\right)+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\Rightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=\frac{3^2-5}{2}=2\)
Ở đâu có 2 thay bằng \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\) là được
a.
\(A=\frac{1}{2\sqrt{3}-2}-\frac{1}{2\sqrt{3}+2}\\ =\frac{2\sqrt{3}+2-2\sqrt{3}+2}{\left(2\sqrt{3}+2\right)\left(2\sqrt{3}-2\right)}\\ =\frac{4}{\left(2\sqrt{3}\right)^2-2^2}=\frac{4}{8}=\frac{1}{2}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}\left(ĐK:x>0;x\ne1\right)\\ =\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b.
\(B=\frac{2}{5}A\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{1}{5}=0\\ \Leftrightarrow\frac{5\left(\sqrt{x}-1\right)-\sqrt{x}}{5\sqrt{x}}=0\\ \Leftrightarrow\frac{4\sqrt{x}-5}{5\sqrt{x}}=0\\ \Rightarrow4\sqrt{x}-5=0\\ \Leftrightarrow\sqrt{x}=\frac{5}{4}\Rightarrow x=\frac{25}{16}\)
(Bạn kiểm tra xem kết quả số có đúng ko nha)
Trả lời:
b, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c, \(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}>\frac{3}{2}\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}>\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Rightarrow2\sqrt{x}+1-3\sqrt{x}>0\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-1\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Vậy \(0< x< 1\) là giá trị cần tìm.
\(a+b+c\le\sqrt{3}\)
\(\Rightarrow ab+bc+ac\le\frac{\left(a+b+c\right)^2}{3}=1\)
Thay vào M ta có: \(M\le\frac{a}{\sqrt{a^2+ab+bc+ac}}+\frac{b}{\sqrt{b^2+ab+bc+ac}}+\frac{c}{\sqrt{c^2+ab+bc+ac}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Xét: \(\left(\frac{a}{a+b}+\frac{a}{a+c}\right)^2\ge\frac{4a^2}{\left(a+b\right)\left(a+c\right)}\Leftrightarrow\frac{a}{a+b}+\frac{a}{a+c}\ge\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Tương tự rồi cộng vế vs vế ta được: \(M\le\frac{\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{a+c}{a+c}}{2}=\frac{3}{2}\)
Dấu = xảy ra khi a=b=c = \(\frac{\sqrt{3}}{3}\)
a) \(ĐKXĐ:-1< a< 1\)
\(B=\left(\frac{1}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}+1\right)\)
\(=\left(\frac{3}{\sqrt{1+a}}+\frac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right):\left[\frac{3}{\sqrt{\left(1-a\right)\left(1+a\right)}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\right]\)
\(=\left[\frac{3}{\sqrt{1+a}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\right]:\frac{3+\sqrt{\left(1+a\right)\left(1-a\right)}}{\sqrt{\left(1+a\right)\left(1-a\right)}}\)
\(=\frac{3+\sqrt{\left(1+a\right)\left(1-a\right)}}{\sqrt{1+a}}.\frac{\sqrt{\left(1+a\right)\left(1-a\right)}}{3+\sqrt{\left(1+a\right)\left(\sqrt{1-a}\right)}}\)
\(=\sqrt{1-a}\)
b) \(a=\frac{\sqrt{3}}{2+\sqrt{3}}\)\(\Rightarrow1-a=1-\frac{\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}=\frac{2}{2+\sqrt{3}}\)
\(=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)
\(=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
Thay \(1-a=\left(\sqrt{3}-1\right)^2\)vào biểu thức ta được:
\(B=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
\(\frac{3}{2}=a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\)
\(\le\frac{a^2+1-b^2}{2}+\frac{b^2+1-c^2}{2}+\frac{c^2+1-a^2}{2}=\frac{3}{2}\)
=> \(\frac{3}{2}\le\frac{3}{2}\)( chỉ xảy ra dấu "=" )
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{cases}}\)=> \(a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)
=> \(B=a^2+b^2+c^2=\frac{3}{2}\)