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:

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

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

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

Hình như tính cái này 

Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)
\(\Rightarrow a^3=\left(\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\right)^3\)
Có (a+b)^3=a^3+b^3+3ab(a+b)
\(\Rightarrow a^3=4+\sqrt{15}+4-\sqrt{15}+3\sqrt[3]{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}a\)
\(\Rightarrow a^3=8+3a\Rightarrow a^3-3a-8=0\)-> khó
 

\(3x=1+\sqrt[3]{\frac{12+3\sqrt[]{15}}{3}}+\sqrt[3]{\frac{12-3\sqrt[]{15}}{3}}\)

\(\Leftrightarrow3x-1=\sqrt[3]{4+\sqrt[]{15}}+\sqrt[3]{4-\sqrt[]{15}}\)

\(\Leftrightarrow\left(3x-1\right)^3=4+\sqrt[]{15}+4-\sqrt[]{45}+3\sqrt[3]{\left(4-\sqrt[]{15}\right)\left(4+\sqrt[]{15}\right)}\left(3x-1\right)\)

<=> \(\left(3x-1\right)^3=8+3\left(3x-1\right)\Leftrightarrow\left(3x-1\right)^3-3\left(3x-1\right)-8=0\)

Khai chuyển rút gọn ta đc 

\(27x^3-27x^2-6=0\Leftrightarrow9x^3-9x^2-2=0\Leftrightarrow9x^3-9x^2-3=-1\)

<=> M  = ( - 1 )^2 = 1 

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{5\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3x}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)

\(=\dfrac{x}{3\sqrt{x}-1}\)

b) Ta có: \(9x^2-10x+1=0\)

\(\Leftrightarrow\left(9x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào P, ta được:

\(P=\dfrac{1}{3-1}=\dfrac{1}{2}\)

c) Thay \(x=8-2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{8-2\sqrt{7}}{3\left(\sqrt{7}-1\right)-1}=\dfrac{8-2\sqrt{7}}{3\sqrt{7}-4}\)

\(=\dfrac{-10+16\sqrt{7}}{47}\)

a)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{3x-2\sqrt{x}-1-3\sqrt{x}+4+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

\(P=\dfrac{x+1}{3\sqrt{x}-1}\)

b) Từ phương trình suy ra \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

Vói x=1 

\(P=\dfrac{1}{3\sqrt{1}-1}=\dfrac{1}{2}\)

Với x= 1/9

\(P=\dfrac{\dfrac{1}{9}}{3\sqrt{\dfrac{1}{9}}-1}\) không có nghiệm

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)

\(M=\sqrt{3}-\sqrt{3}+1\)

\(M=1\)

b) Ta có:

\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)

\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Theo đề ta có: \(M=2N\)

Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)

\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)

\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)

\(\Leftrightarrow\sqrt{a}=2\)

\(\Leftrightarrow a=4\left(tm\right)\)

Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

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

\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)

\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)

\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{3x+3\sqrt{x}-1}{9x-1}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)