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=\(\left(3\sqrt{3}-3\sqrt{3}+2\sqrt{6}\right):3\sqrt{3}\)
\(=1-\dfrac{\sqrt{6}}{2}+\dfrac{2\sqrt{2}}{3}\)
=\(\dfrac{6}{6}-\dfrac{3\sqrt{6}}{6}+\dfrac{4\sqrt{2}}{6}\)
=\(\dfrac{6+\sqrt{6}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{36}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{4}{y}=\dfrac{5}{9}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{18}\\\dfrac{1}{x}=\dfrac{1}{36}-\dfrac{1}{18}=-\dfrac{1}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=18\\x=-36\end{matrix}\right.\)
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ĐKXĐ: \(5x^2+2x-3>=0\)
=>\(5x^2+5x-3x-3>=0\)
=>\(\left(x+1\right)\left(5x-3\right)>=0\)
TH1: \(\left\{{}\begin{matrix}x+1>=0\\5x-3>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-1\\x>=\dfrac{3}{5}\end{matrix}\right.\)
=>\(x>=\dfrac{3}{5}\)
TH2: \(\left\{{}\begin{matrix}x+1< =0\\5x-3< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =-1\\x< =\dfrac{3}{5}\end{matrix}\right.\)
=>\(x< =-1\)
\(\left(x+1\right)\cdot\sqrt{5x^2+2x-3}=5x^2+4x-5\)
=>\(\left(x+1\right)\sqrt{5x^2+2x-3}=5x^2+2x-3+2x-2\)
=>\(\left(x+1\right)\sqrt{5x^2+2x-3}-\left(5x^2+2x-3\right)-\left(2x-2\right)=0\)
=>\(\sqrt{5x^2+2x-3}\left(x+1-\sqrt{5x^2+2x-3}\right)-2\left(x-1\right)=0\)
=>\(\sqrt{5x^2+2x-3}\cdot\dfrac{\left(x+1\right)^2-\left(5x^2+2x-3\right)}{x+1+\sqrt{5x^2+2x-3}}-2\left(x-1\right)=0\)
=>\(\sqrt{5x^2+2x-3}\cdot\dfrac{x^2+2x+1-5x^2-2x+3}{x+1+\sqrt{5x^2+2x-3}}-2\left(x-1\right)=0\)
=>\(\dfrac{\sqrt{5x^2+2x-3}}{x+1+\sqrt{5x^2-2x+3}}\cdot\left(-4x^2+4\right)-2\left(x-1\right)=0\)
=>\(\dfrac{2\sqrt{5x^2+2x-3}}{x+1+\sqrt{5x^2-2x+3}}\cdot\left(x^2-1\right)+\left(x-1\right)=0\)
=>\(\dfrac{2\sqrt{5x^2+2x-3}\cdot\left(x+1\right)\left(x-1\right)}{x+1+\sqrt{5x^2-2x+3}}+\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(\dfrac{2\sqrt{5x^2+2x-3}\cdot\left(x+1\right)}{x+1+\sqrt{5x^2-2x+3}}+1\right)=0\)
=>x-1=0
=>x=1(nhận)
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Với \(n\in N;n>0\) có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Áp dụng vào P có:
\(P=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2016}}-\dfrac{1}{\sqrt{2017}}\)
\(=1-\dfrac{1}{\sqrt{2017}}\)
\(\Rightarrow a^2+b=1^2+2017=2018\)
Ý A
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1: \(=\left[\left(\sqrt{2}+\sqrt{3}\right)^2-5\right]\cdot\left[\left(\sqrt{5}\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2\right]\)
\(=2\sqrt{6}\left(5-5+2\sqrt{6}\right)=2\sqrt{6}\cdot2\sqrt{6}=24\)
2: \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=>\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
=>\(A=\sqrt{5}+1\)
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\(f\left(a,b\right)=a^2+8b^2-6ab+14a-40b+48=3\)
\(\Leftrightarrow f\left(a,b\right)=a^2+8b^2-6ab+14a-40b+45=0\)
\(\Leftrightarrow a^2+2a\left(7-3b\right)+\left(8b^2-40b+45\right)=0\)
Xét \(\Delta'=\left(7-3b\right)^2-\left(8b^2-40b+45\right)=b^2-2b+4=\left(b-1\right)^2+3>0\)
Vậy PT luôn có hai nghiệm phân biệt.
Vì a,b nguyên nên \(b^2-2b+4=k^2\left(k\in N\right)\)
\(\Leftrightarrow k^2-\left(b-1\right)^2=3\Leftrightarrow\left(k-b+1\right)\left(k+b-1\right)=3\)
Xét các trường hợp với k-b+1 và k+b-1 là các số nguyên được :
(b;k) = (0;2) ; (0;-2) ; (2;2) ; (2;-2)
Thay lần lượt các giá trị của b vào f(a,b) = 3 để tìm a.
Vậy : (a;b) = (-9;0) ; (-5;0) ; (-3;2) ; (1;2)
a, Thay x = 36 vào B ta được : \(B=\frac{6}{6-3}=\frac{6}{3}=2\)
b, \(B< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{1}{2}< 0\Leftrightarrow\frac{\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)Với \(x>0;x\ne9\)
\(\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\Rightarrow0< x< 9\)
c, Với \(x>0;x\ne1\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\left(\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
d, \(P=AB=\frac{\sqrt{x}+2}{\sqrt{x}-3}=\frac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\frac{5}{\sqrt{x}-3}\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)