\(\sqrt{9}\)
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Đặt \(B=\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}\)
=>\(B^2=9+\sqrt{5}+9-\sqrt{5}+2\cdot\sqrt{81-5}\)
=>\(B^2=18+2\sqrt{76}\)
=>\(B=\sqrt{18+2\sqrt{76}}\)
\(\Leftrightarrow A=\sqrt{\dfrac{18+2\sqrt{76}}{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{\dfrac{2\left(9+2\sqrt{19}\right)}{9+2\sqrt{19}}}-\sqrt{2}+1=\sqrt{2}-\sqrt{2}+1=1\)
\(=9\sqrt{x}-9.\dfrac{1}{3}.\sqrt{x}+x.\dfrac{1}{\sqrt{x}}.\sqrt{9}-3x\)
\(=9\sqrt{x}-3\sqrt{x}+3\sqrt{x}-3x\)
\(=-3x+9\sqrt{x}\)
\(=9\sqrt{x}-9\cdot\dfrac{1}{3}\sqrt{x}+3\sqrt{\dfrac{x^2}{x}}-x\sqrt{9}\\ =9\sqrt{x}-3\sqrt{x}+3\sqrt{x}-3x\\ =9\sqrt{x}-3x=3\sqrt{x}\left(3\sqrt{x}-1\right)\)
Ta có:
\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}\)
\(=\frac{\sqrt{n+4}-\sqrt{n}}{4\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)
Áp dụng vào bài toán ta được
\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)
\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)
Không hiểu sao cứ gửi ảnh nó lại bị lộn xộn nên bạn cố nhìn nhé
( ͡°( ͡° ͜ʖ( ͡° ͜ʖ ͡°)ʖ ͡°) ͡°)
ĐKXĐ: \(x,y\ge0\)
Từ hệ pt trên suy ra \(\sqrt{x}+\sqrt{y+9}=\sqrt{y}+\sqrt{x+9}\Leftrightarrow x+y+9+2\sqrt{xy+9x}=x+y+9+2\sqrt{yx+9y}\Leftrightarrow2\sqrt{xy+9x}=9+2\sqrt{xy+9y}\Leftrightarrow x=y\)
Vậy hệ pt thành \(\sqrt{x}+\sqrt{x+9}=9\Leftrightarrow2x+9+2\sqrt{x^2+9x}=81\Leftrightarrow2\sqrt{x^2+9x}=72-2x\)
( đk : x \(\le\) 36, y \(\le36\))
\(\Leftrightarrow4\left(x^2+9x\right)=5184-288x+4x^2\Leftrightarrow324x=5184\Leftrightarrow x=16\) (tmđk)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(c,\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\\ =\sqrt{\sqrt{3^2}+2\sqrt{3}.1+1}+\sqrt{\sqrt{3^2}-2\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(d,\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\\ =\sqrt{\sqrt{5^2}+2.2\sqrt{5}+2^2}-\sqrt{\sqrt{5^2}-2.2\sqrt{5} +2^2}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)
\(\sqrt{9+2\sqrt{20}}-\sqrt{9-2\sqrt{20}}=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|=\sqrt{5}+2-\sqrt{5}+2=4\)
= \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=2+\sqrt{5}+2-\sqrt{5}=4\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
=-4
7x7=49
= 14
nho ket ben voi minh nhe