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A=\(\frac{u-v}{\sqrt{u}+\sqrt{v}}-\frac{\sqrt{u^3}+\sqrt{v^3}}{u-v}=\frac{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}{\sqrt{u}+\sqrt{v}}-\frac{\left(\sqrt{u}+\sqrt{v}\right)\left(u-\sqrt{u}\sqrt{v}+v\right)}{\left(\sqrt{u}+\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\frac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}=\frac{u-2\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}=\frac{-\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)
\(A=\dfrac{u-v}{\sqrt{u}+\sqrt{v}}-\dfrac{\sqrt{u^3}+\sqrt{v^3}}{u-v}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u\sqrt{u}+v\sqrt{v}}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)
\(=\dfrac{\left(\sqrt{u}-\sqrt{v}\right)\sqrt{u}-\left(\sqrt{u}-\sqrt[]{v}\right)\sqrt{v}-\left(u-\sqrt{uv}+v\right)}{\sqrt{u}-\sqrt{v}}\)
\(=\dfrac{u-\sqrt{uv}-\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}\)
\(\Leftrightarrow\)\(-\dfrac{\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)
\(B=\dfrac{2u+\sqrt{uv}-3v}{2u-5\sqrt{uv}+3v}\)
\(=\dfrac{2u+3\sqrt{uv}-2\sqrt{uv}-3v}{2u-2\sqrt{uv}-3\sqrt{uv}+3v}\)
\(=\dfrac{\sqrt{u}.\left(2\sqrt{u}+3\sqrt{v}\right)-\sqrt{v}.\left(2\sqrt{u}+3\sqrt{v}\right)}{2\sqrt{u}.\left(\sqrt{u}-\sqrt{v}\right)-3\sqrt{v}.\left(\sqrt{u}-\sqrt{v}\right)}\)
\(=\dfrac{\left(2\sqrt{u}+3\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}{\left(\sqrt{u}-\sqrt{v}\right)\left(2\sqrt{u}-3\sqrt{v}\right)}\)
\(=\dfrac{2\sqrt{u}+3\sqrt{v}}{2\sqrt{u}-3\sqrt{v}}\\ =\dfrac{4u+12\sqrt{uv}+9v}{4u-9v}\)
\(M=\frac{2\sqrt{y}}{x-y}+\frac{\sqrt{x}+\sqrt{y}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{x-y}=\frac{2\sqrt{y}+\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{x-y}=\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{2}{\sqrt{x}-\sqrt{y}}\)
b/ Khi \(x=4y\) và M=1
\(\Leftrightarrow\frac{2}{\sqrt{4y}-\sqrt{y}}=1\Leftrightarrow\frac{2}{2\sqrt{y}-\sqrt{y}}=1\Leftrightarrow\frac{2}{\sqrt{y}}=1\)
\(\Leftrightarrow\sqrt{y}=2\Rightarrow y=4\Rightarrow x=16\)
mình có sửa lại đề 1 chút!
đặt \(T=\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)
đặt \(u=a^4;v=b^6\)(a,b>0) ta có
\(T=\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}=\frac{a^4-8a^2b^2+4b^2}{a^2-2b^2+2ab}+3b^2\)
vậy \(T=\frac{a^4-8a^2b^2+4b^4}{a^2-2b^2+2ab}+3b^2=\frac{a^4-5a^2b^2-2b^4+6ab^3}{a^2-2b^2+2ab}=a^2-2ab+b^2\)
từ đó suy ra \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}\)
vì \(u^3\ge v^2\)nên \(\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}=\sqrt[4]{u}\)
\(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)
với u=1 ta có \(T=\sqrt{\frac{1-8\sqrt[6]{v^2}+4\sqrt[3]{v^2}}{1-2\sqrt[3]{v}+2\sqrt[6]{v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}\)
nếu \(1-2\sqrt[3]{v}+2\sqrt[6]{v}=0\)thì \(\sqrt[3]{v}=\frac{3+1}{2}>0\)
do \(v^2>1=u^3\), mâu thuẫn suy ra \(1-2\sqrt[3]{v}+2\sqrt[6]{v}\ne0\)
tóm lại với \(u^3\ge v^2\)và u,v\(\inℚ^+\)để \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)cần và đủ là u=1 và v<1, v\(\inℚ^+\)được lấy tùy ý
\( a)A = \dfrac{{a - \sqrt a - 6}}{{4 - a}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{a + 2\sqrt a - 3\sqrt a - 6}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 3}}{{\sqrt a - 2}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 2}}{{\sqrt a - 2}} = - 1 \)
\( b)B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}\\ B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{\sqrt x + 1}} \)
\(\left[\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}\right]\left[\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]^2=\left(x+\sqrt{x}+1\right)\frac{1}{\left(1+\sqrt{x}\right)^2}=\frac{x+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
Đề bài sai
\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(\sqrt{2011}-\sqrt{2010}=\frac{1}{\sqrt{2011}+\sqrt{2010}}\)
Do \(\sqrt{2012}>\sqrt{2010}\) \(\Rightarrow\sqrt{2012}+\sqrt{2011}>\sqrt{2011}+\sqrt{2010}>0\)
\(\Rightarrow\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\Rightarrow\sqrt{2012}-\sqrt{2011}< \sqrt{2011}-\sqrt{2010}\)
\(A=\frac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
\(M^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\le2\left(x-1+9-x\right)=16\)
\(\Rightarrow M\le4\Rightarrow M_{max}=4\) khi \(x-1=9-x\Leftrightarrow x=5\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{4+4\sqrt{3}+3}\)
\(=2-\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}=2-\sqrt{3}+\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(a-2\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)
\(=\left|2-\sqrt{3}\right|+\sqrt{3+4\sqrt{3}+4}\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=2-\sqrt{3}+\left|\sqrt{3}+2\right|\)
\(=2-\sqrt{3}+\sqrt{3}+2\)
\(=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)( \(\hept{\begin{cases}a,b\ge0\\a\ne b\end{cases}}\))
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-2\sqrt{ab}+b\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-2\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-b}{a-b}=1\)
\(ĐKXĐ:x\ne9,x\ge0\)
Ta có : \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
Ta thấy : \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3>0\)
\(\Rightarrow\frac{3}{\sqrt{x}+3}\le\frac{3}{3}=1\)
Hay : \(A\le1\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy GTLN của \(A=1\) khi \(x=0\)
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Câu hỏi của Duy Saker Hy - Toán lớp 9 - Học toán với OnlineMath