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chịu thua vô điều kiện xin lỗi nha : v
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
c/
\(\left(x-4\right)P+y^2+2xy+1+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x^2-1\right)}{x-4}+y^2+2xy+1+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow x^2+y^2+2xy+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left|2x+3y+1\right|=0\)
Do \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left|2x+3y+1\right|\ge0\end{matrix}\right.\) \(\forall x;y\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2x+3y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+3}+\frac{x^2-x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\right)\)
\(P=\left(\frac{x-4+x^2-x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{x+3\sqrt{x}+\sqrt{x}+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\right)\)
\(P=\left(\frac{x^2-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2}\right)\)
\(P=\frac{x^2-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}.\left(\frac{\sqrt{x}+3}{\sqrt{x}+2}\right)\)
\(P=\frac{x^2-1}{x-4}\)
b/ Để \(P\ge0\Leftrightarrow\frac{x^2-1}{x-4}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1\ge0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1\le0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>4\\-1\le x\le1\end{matrix}\right.\)
Kết hợp với ĐKXĐ \(x\ge0\), \(\Leftrightarrow\left[{}\begin{matrix}x>4\\0\le x\le1\end{matrix}\right.\)