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\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
\(b1:=\sqrt{2}\left(\sqrt{3}+1\right).\sqrt{2-\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\sqrt{4-2\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)\\ =2\\ \\ b2:a,=\sqrt{\dfrac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}-3\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{2}}{\sqrt{2}}.\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{54-14\sqrt{5}}}{2\sqrt{10}-3\sqrt{2}} .\left(\sqrt{10}-\sqrt{2}\right)\\ \)\(=\dfrac{\sqrt{\left(7-\sqrt{5}\right)^2}}{2\sqrt{10}-3\sqrt{2}}.\left(\sqrt{10}-\sqrt{2}\right)\)\(\\ =\dfrac{8\sqrt{10}-12\sqrt{2}}{2\sqrt{10}-3\sqrt{2}}\\ =4\)
\(a,A=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left(\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(b,A=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
Để A nguyên thì \(\sqrt{x}+3\inƯ\left(2\right)\)
\(\Rightarrow\sqrt{x}+3\in\left\{1;2\right\}\) ( vì \(x\ge0\) )
Với \(\sqrt{x}+3=1\)\(\Rightarrow\sqrt{x}=-2\) ( loại vì \(\sqrt{x}\ge0\) )
Với \(\sqrt{x}+3=2\) \(\Rightarrow\sqrt{x}=-1\) ( loại )
=> ......
a ) Ngại làm quá >,,<
Ơ nhưng mà phân thức \(\dfrac{x+3}{x-9}\) đáng nhẽ phải là \(\dfrac{\sqrt{x}+3}{x-9}\) chứ nhỉ ???
\(1.\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=|\sqrt{x-1}-1|=\sqrt{x-1}-1\)
\(2.\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(2x-3\right)^2}}=\dfrac{1}{|2x-3|}\)
\(3.\dfrac{1}{\sqrt{x+2\sqrt{x-1}}}=\dfrac{1}{\sqrt{x-1+2\sqrt{x-1}+1}}=\dfrac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}=\dfrac{1}{|\sqrt{x-1}+1|}\)
bạn cho mình hỏi tại sao từ \(\sqrt{x-1-2\sqrt{x-1}+1}\) sang \(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)
Vậy $m=2$
Rút gọn P:
\(P=\dfrac{4x}{\sqrt{x}-3}\)
\(\Rightarrow m\left(\sqrt{x}-3\right)P>x+1\)
\(\Leftrightarrow4mx>x+1\)
\(\Leftrightarrow\left(4m-1\right)x>1\)(1)
Xét \(4m-1=0\)
\(\Leftrightarrow m=\dfrac{1}{4}\)(loại vì (1) sai)
\(\Leftrightarrow x>\dfrac{1}{4m-1}\)
Xét \(\dfrac{1}{4m-1}< 0\)
\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x>9\end{matrix}\right.\)(loại)
Xét \(\Leftrightarrow\dfrac{1}{4m-1}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m-1>0\\\dfrac{1}{4m-1}\le9\end{matrix}\right.\)
\(\Leftrightarrow m\ge\dfrac{5}{18}\)
Có mấy cái dâu tương đương. Copy từ cái hàng trên xuống dưới quên xóa nha. Mà chắc đọc hiểu thôi ah.
Mọi người ơi, giúp em với ạ!
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)