Rút gọn biểu thức B
B = (\(\dfrac{15-\sqrt{x}}{x-25}\) + \(\dfrac{2}{\sqrt{x}+5}\)) : \(\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\)
K
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17 tháng 9 2021
\(a,B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\left(x\ge0;x\ne1\right)\\ B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
17 tháng 9 2021
b: Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)
\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
NU
28 tháng 8 2021
\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
9 tháng 9 2021
\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
9 tháng 9 2021
\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)
23 tháng 9 2021
\(C=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(đk:x\ge0,x\ne25\right)\)
\(=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)
23 tháng 9 2021
\(ĐK:x\ge0;x\ne25\)
\(C=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ C=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)
12 tháng 9 2021
Ta có: \(P=\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(=\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
T
27 tháng 9 2023
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\left(x\ge0;x\ne25\right)\)
Để \(A=\dfrac{2\sqrt{x}}{3}\) thì:
\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{2\sqrt{x}}{3}\)
\(\Leftrightarrow3\sqrt{x}-15=2x+10\sqrt{x}\)
\(\Leftrightarrow2x+10\sqrt{x}-3\sqrt{x}+15=0\)
\(\Leftrightarrow2x+7\sqrt{x}+15=0\)
Mà \(2x+7\sqrt{x}+15>0\) (vì \(x\ge0\))
nên không tìm được giá trị nào của \(x\) thoả mãn \(A=\dfrac{2\sqrt{x}}{3}\)
#\(Toru\)
17 tháng 5 2022
\(A=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)
24 tháng 8 2021
Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5}{\sqrt{x}+3}\)
\(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\left(đk:x\ne25,x\ge0\right)\)
\(=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+3\right)}=\dfrac{1}{\sqrt{x}+3}\)
Ta có: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\)
\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\cdot\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(=\dfrac{1}{\sqrt{x}+3}\)