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Từ \(x=\frac{1}{3}\left(1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)

\(\Rightarrow3x-1=\left(\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)

\(\Leftrightarrow\left(3x-1\right)^3=\left(\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)^3\)

\(\Rightarrow\left(3x-1\right)^3=8+3\left(3x+1\right)\)

\(\Leftrightarrow9x^3-9x^2-2=0\)

\(\Rightarrow M=-1\)

Do \(12=\sqrt{144}>\sqrt{135}\) nên \(x>0\)

Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)

\(a^3=8+3\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)=8+3a\)

Ta có: \(x=\dfrac{1}{3}\left(a+1\right)\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\)

Lại có: \(x^3=\dfrac{1}{27}\left(a+1\right)^3\Leftrightarrow9x^3=\dfrac{1}{3}\left(a^3+3a^2+3a+1\right)\)

\(\Leftrightarrow9x^3=\dfrac{1}{3}\left(8+3a+3a^2+3a+1\right)=a^2+2a+3\)

\(\Rightarrow M=\left(a^2+2a+3-a^2-2a-1-3\right)^2=\left(-1\right)^2=1\)

Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)

\(\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\) (1)

\(x^3=\dfrac{1}{27}\left(a+1\right)^3=\dfrac{1}{27}\left(a^3+3a^2+3a+1\right)\)

Ta có:

\(a^3=\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)^3\)

\(\Rightarrow a^3=\dfrac{24}{3}+3\sqrt[3]{\dfrac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{9}}.\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)\)

\(\Rightarrow a^3=8+3a\)

\(\Rightarrow x^3=\dfrac{1}{27}\left(8+3a+3a^2+3a+1\right)=\dfrac{1}{9}\left(a^2+2a+3\right)\)

\(\Rightarrow9x^3=a^2+2a+3\) (2)

Thay (1), (2) vào M ta được:

\(M=\left(9x^3-9x^2-3\right)^2=\left(a^2+2a+3-\left(a^2+2a+1\right)-3\right)^2\)

\(\Rightarrow M=\left(-1\right)^2=1\)

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) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)

2)a) \(P=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

\(=\frac{a-2\sqrt{a}+1}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}-1\)

b) \(19-8\sqrt{3}=\left(\sqrt{3}-4\right)^2\Rightarrow P=\sqrt{\left(\sqrt{3}-4\right)^2}-1=4-\sqrt{3}-1=3-\sqrt{3}\)

c) P < 1 <=> \(\sqrt{a}-1< 1\Leftrightarrow a< 4\)

Kết hợp với điều kiện : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)

mk làm luôn.

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

=\(\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

\(\frac{3.\left(x+\sqrt{x}\right).\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^

a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)

\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)

b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)

\(\Rightarrow9x+7\sqrt{x}-19=0\)

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