\(A=\frac{x-21}{x-6\sqrt{x}+5}+\frac{1}{\sqrt{x}—1}+\frac{1}{5-\sqrt{x}}\)
Rút gọn A
Giúp mình vs mình cần gấp lắm\(\)
K
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NC
6 tháng 8 2020
Ta có: \(B=\frac{\sqrt{\frac{1}{9}}-3}{\sqrt{\frac{1}{9}}-1}\)
\(B=\frac{\frac{1}{3}-3}{\frac{1}{3}-1}\)
\(B=\frac{-\frac{8}{3}}{-\frac{2}{3}}=4\)
NC
6 tháng 8 2020
đkxđ: \(\hept{\begin{cases}x\ne1\\x\ne25\end{cases}}\)
Ta có:
\(A=\frac{x-21}{x-6\sqrt{x}+5}+\frac{1}{\sqrt{x}-1}+\frac{1}{5-\sqrt{x}}\)
\(A=\frac{x-21}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-5}\)
\(A=\frac{x-21+\sqrt{x}-5-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{x-25}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{\sqrt{x}+5}{\sqrt{x}-1}\)
NC
13 tháng 8 2020
Bài làm:
đkxđ: \(x\ne4;x\ne9\)
Ta có:
\(P=\frac{2\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(P=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(P=\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{2\sqrt{x}-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
NN
13 tháng 8 2020
\(ĐKXĐ:4< x< 9\)
\(P=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(2\sqrt{x}-9\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
19 tháng 7 2017
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
4 tháng 10 2020
\(A=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
ĐKXĐ : x > 1
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}+\frac{1}{\sqrt{x}-1}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\times\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{x}{\sqrt{x}-1}\)
Để A = 9/2
=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )
<=> 2x = 9( √x - 1 )
<=> 2x = 9√x - 9
<=> 2x + 9 = 9√x (1)
Bình phương hai vế
(1) <=> 4x2 + 36x + 81 = 81x
<=> 4x2 + 36x + 81 - 81x = 0
<=> 4x2 - 45x + 81 = 0
<=> 4x2 - 36x - 9x + 81 = 0
<=> 4x( x - 9 ) - 9( x - 9 ) = 0
<=> ( x - 9 )( 4x - 9 ) = 0
<=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )
20 tháng 9 2018
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
3 tháng 9 2020
Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) ( ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))
\(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)
\(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)
11 tháng 10 2016
Bạn tự tìm ĐKXĐ nhé :)
Xét tử thức : \(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
Xét mẫu thức : \(\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}=\sqrt{\left(\frac{4}{x}-1\right)^2}=\left|\frac{4}{x}-1\right|=\left|\frac{x-4}{x}\right|\)
Từ đó rút gọn P
ĐKXĐ: x>=0; x khác 1; x khác 25.
\(A=\frac{x-21}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-5}.\)
=\(\frac{x-21}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}+\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}\)
\(=\frac{x-21+\sqrt{x}-5-\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}=\frac{x-25}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-5\right)}.\)
\(=\frac{\sqrt{x}+5}{\sqrt{x}-1}.\)
Kết luận: ...