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b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
a: \(A=\left(\dfrac{\sqrt{x}}{x+2}+\dfrac{6\sqrt{x}}{x-4}\right)\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{x-2\sqrt{x}+6\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}+2}{1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\)
b: \(M=A:B=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}+1}\)
b: \(M-1=\dfrac{x+4\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{x+3\sqrt{x}-1}{\sqrt{x}+1}>0\)
=>M>1
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
a: \(A=\left(\dfrac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{1}{\sqrt{a}-1}\right):\dfrac{a+\sqrt{a}+1-a-4}{a+\sqrt{a}+1}\)
\(=\dfrac{2a+1-a-\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\cdot\dfrac{a+\sqrt{a}+1}{\sqrt{a}-3}\)
\(=\dfrac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-3}\)
b: Để A là số nguyên dương thì \(\left\{{}\begin{matrix}a>9\\\sqrt{a}-3+3⋮\sqrt{a}-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>9\\\sqrt{a}-3\in\left\{1;-1;3;-3\right\}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a>9\\a\in\left\{16;36;4;0\right\}\end{matrix}\right.\Leftrightarrow a\in\left\{16;36\right\}\)
\(1.ĐKXĐ\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
\(M=\frac{a+1}{\sqrt{a}}+\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
2.
có \(M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
vì \(a>0\) và \(a\ne0\)
\(\Rightarrow\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\Leftrightarrow4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>4\)
Vậy M > 4
3.
M > 4\(\Leftrightarrow0\le N< \frac{3}{2}=1,5\)
Để N nhận giá trị nguyên
\(\Rightarrow N=1\)
\(\Rightarrow\frac{6}{M}=1\Leftrightarrow M=6\)
\(\Rightarrow\frac{a+2\sqrt{a}+1}{\sqrt{a}}=6\Leftrightarrow a+2\sqrt{a}+1=6\sqrt{a}\)
\(\Leftrightarrow a-4\sqrt{a}+1=0\)
Có \(\Delta`=2+1=3>0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=2+\sqrt{3}\left(tm\right)\\\sqrt{a}=2-\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a=7+4\sqrt{3}\)
Vậy \(a=7+4\sqrt{3}\) thì N nhận giá trị nguyên
