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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) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)