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\(\dfrac{x+2\sqrt{x}}{\sqrt{x}-1}=8\left(x\ge0;x\ne1\right)\)
\(\Leftrightarrow x+2\sqrt{x}=8\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+2\sqrt{x}=8\sqrt{x}-8\)
\(\Leftrightarrow x+2\sqrt{x}-8\sqrt{x}+8=0\)
\(\Leftrightarrow x-6\sqrt{x}+8=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\left(tm\right)\)
Vậy: ...
\(\dfrac{x+2\sqrt{x}}{\sqrt{x}-1}=8\left(x\ge0,x\ne1\right)\\ < =>x+2\sqrt{x}=8\sqrt{x}-8\\ < =>x-6\sqrt{x}+8=0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\left(TMDK\right)\)
\(=>S=\left\{4;16\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+\frac{2}{\sqrt{x}+1};B=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{x+\sqrt{x}-2}\)
đề bài là thế này ạ!?
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\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
\(a,\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) \(\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\\ =\sqrt{x}\)
\(b,\dfrac{1-2\sqrt{x}+x}{1-\sqrt{x}}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\dfrac{1^2-2\sqrt{x}+\sqrt{x^2}}{1-\sqrt{x}}\\ =\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}\\ =1\)