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Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)
Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì \(\left(a-b=b-c\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)
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Ta có: \(\left(\sqrt{a}+\sqrt{c}\right)^2=a+2\sqrt{ac}+c=2b+2\sqrt{ac}\)(1)
Lại có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}\)
\(=\frac{\left(2\sqrt{b}+\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(Nhân cả tử & mẫu với \(\sqrt{a}+\sqrt{c}\))
\(=\frac{2\sqrt{ab}+2\sqrt{bc}+\left(\sqrt{a}+\sqrt{c}\right)^2}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(2)
Thế (1) và (2) => \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}\)\(=\frac{2\sqrt{ab}+2\sqrt{bc}+2b+\sqrt{ca}}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}=\frac{2\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\frac{2}{\sqrt{a}+\sqrt{c}}.\)
\(\Rightarrow\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)(đpcm).
\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=4\sqrt{2}\)
\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|1-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
Lời giải:
a.
\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)
b.
\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
c.
\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)
a ĐKXĐ \(a\ge0,a\ne\dfrac{1}{4},a\ne1\)
\(\Rightarrow P=1+\left(\dfrac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
= \(1+\left(\dfrac{\left(-1\right)\left(2\sqrt{a}-1\right)}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{2\sqrt{a}-1}\)
= \(1+\left(-1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a+\sqrt{a}+1}\right)\sqrt{a}\)
= \(1-\sqrt{a}+\dfrac{a\sqrt{a}+a}{a+\sqrt{a}+1}\) = \(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{1-a\sqrt{a}+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{a+1}{a+\sqrt{a}+1}\)
b Xét hiệu \(P-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}=\dfrac{3a+3-2a-2\sqrt{a}-2}{a+\sqrt{a}+1}=\dfrac{a-2\sqrt{a}+1}{a+\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)^2}{a+\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(\Rightarrow P>\dfrac{2}{3}\)
c Ta có \(P=\dfrac{\sqrt{6}}{\sqrt{6}+1}\Rightarrow\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{\sqrt{6}+1}\) \(\Rightarrow\left(a+1\right)\left(\sqrt{6}+1\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\Leftrightarrow a\sqrt{6}+a+\sqrt{6}+1=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)
\(\Leftrightarrow a-\sqrt{6a}+1=0\Leftrightarrow a-\sqrt{6a}+\dfrac{6}{4}-\dfrac{2}{4}=0\Leftrightarrow\left(\sqrt{a}-\dfrac{\sqrt{6}}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{\sqrt{6}+1}{2}\\\sqrt{a}=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\) (Do \(\sqrt{a}\ge0\)) \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\)
Vậy...
\(P=\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(3+2\sqrt{2.3}+2\right)\sqrt{3-2\sqrt{3.2}+2}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}+\sqrt{2}}\)
=\(\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
=\(3-2=1\)
ta có : \(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\ge2\sqrt{\dfrac{1}{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}+b}}\)
\(\ge\dfrac{2}{\sqrt{a+b+c+b}}=\dfrac{2}{\sqrt{4b}}=\dfrac{2}{2\sqrt{b}}=\dfrac{1}{\sqrt{b}}=\dfrac{2}{\sqrt{a+c}}\ge\dfrac{2}{\sqrt{a}+\sqrt{b}}\)
dấu "=" xảy ra khi \(a=b=c\Leftrightarrow a+c=2b\Rightarrow\left(đpcm\right)\)