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\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b1:=\sqrt{2}\left(\sqrt{3}+1\right).\sqrt{2-\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\sqrt{4-2\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)\\ =2\\ \\ b2:a,=\sqrt{\dfrac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}-3\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{2}}{\sqrt{2}}.\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{54-14\sqrt{5}}}{2\sqrt{10}-3\sqrt{2}} .\left(\sqrt{10}-\sqrt{2}\right)\\ \)\(=\dfrac{\sqrt{\left(7-\sqrt{5}\right)^2}}{2\sqrt{10}-3\sqrt{2}}.\left(\sqrt{10}-\sqrt{2}\right)\)\(\\ =\dfrac{8\sqrt{10}-12\sqrt{2}}{2\sqrt{10}-3\sqrt{2}}\\ =4\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-1}{\sqrt{x}+1}\)
Thay x=3 vào B, ta được:
\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a/
\(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{\left(3\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3x+6\sqrt{x}+3}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x+2}{\sqrt{x}}\)
Nếu không phiền, bạn có thể giải chi tiết cho mình được không ạ. Mình cảm ơn nhiều !
\(a,A=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left(\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(b,A=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
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 ???
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.
Bo thi:>
+ đk x > 0 , x khác 1