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a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b) đề sai rồi nha
c) \(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}=\dfrac{a\sqrt{a}-4\sqrt{a}+2a-8}{a-4}\)
\(=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)\left(a-4\right)}{a-4}=\sqrt{a}+2\)
a) ...= \(\dfrac{1}{4}\).\(6\sqrt{5}\) +\(2\sqrt{5}\) - \(3\sqrt{5}\) +5
= \(\dfrac{3}{2}\sqrt{5}\) -\(\sqrt{5}\) +5
=5 - \(\dfrac{1}{2}\sqrt{5}\)
d) ...= \(\sqrt{\dfrac{a}{\left(1+b\right)^2}}\) . \(\sqrt{\dfrac{4a\left(1+b\right)^2}{15^2}}\)
= \(\sqrt{\dfrac{4a^2\left(1+b\right)^2}{\left(1+b\right)^2.15^2}}\) = \(\sqrt{\dfrac{4a^2}{15^2}}\)= \(\dfrac{2a}{15}\)
\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )
\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )
a) \(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2\cdot\left(x+y\right)\cdot\sqrt{3}}{\left(x+y\right)\cdot\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{3}}{\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{6}}{2\left(x-y\right)}=\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left[\left(2a\right)^2-2\cdot2\cdot a+1^2\right]}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left(2a-1\right)^2}=\dfrac{2}{2a-1}\cdot a\cdot\left(2a-1\right)\cdot\sqrt{5}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)
Lời giải:
\(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow ax=\sqrt{\frac{2a-b}{b}}\)
\(\Rightarrow 1+ax=\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}; 1-ax=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)
\(\Rightarrow \frac{1-ax}{1+ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\frac{(\sqrt{b}-\sqrt{2a-b})^2}{2(b-a)}\)
Lại có:
\(\frac{1+bx}{1-bx}=\frac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\frac{a^2-(2ab-b^2)}{(a-\sqrt{2ab-b^2})^2}=\frac{(a-b)^2}{(a-\sqrt{2ab-b^2})^2}\)
\(\Rightarrow \sqrt{\frac{1+bx}{1-bx}}=\frac{b-a}{a-\sqrt{2ab-b^2}}\)
Do đó:
$A=\frac{(\sqrt{b}-\sqrt{2a-b})^2}{2a-2\sqrt{2ab-b^2}}=\frac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1$
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)
Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)
\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)
\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)