\(P=
\frac{3x++\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)
Rút gọn P
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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
\(ĐKXĐ:4< x< 9\)
\(P=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(2\sqrt{x}-9\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
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 )
24 tháng 7 2020
Bài 1:
Ta có: \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=3^2\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của PT \(S=\left\{11\right\}\)
PN
6 tháng 8 2020
Bài 1
a, Với \(x=9\)thì \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{3}{3}+1=2\)
b, Để \(A=\frac{5}{2}\)thì \(\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{5}{2}< =>\frac{3}{\sqrt{x}}=\frac{3}{2}< =>x=4\)
Bài 2
a, \(B=\frac{\sqrt{x}-2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}\left(đk:x>0\right)\)
\(=1-\frac{2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}=\frac{x+5\sqrt{x}+2}{x+\sqrt{x}}-\frac{2}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}+5x+2\sqrt{x}-2x-2\sqrt{x}}{x\sqrt{x}+x}=\frac{x\sqrt{x}+3x}{x\sqrt{x}+x}\)
\(=1+\frac{2x}{x\left(\sqrt{x}+1\right)}=1+\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
A
6 tháng 8 2020
\(A=\frac{3+\sqrt{x}}{\sqrt{x}}\)Thay x = 9 ta có :
\(VT=\frac{3+\sqrt{9}}{\sqrt{9}}=\frac{3+3}{3}=2\)
Bài ra ta có : \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\Leftrightarrow\frac{3}{\sqrt{x}}+1=\frac{5}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}}=\frac{3}{2}\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
\(P=\frac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+\sqrt{x}-2}\)
\(P=\frac{3x+3\sqrt{x}-3-x+1-x+4}{x+\sqrt{x}-2}\)
\(P=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)