\(\left(3x-1\right)^3=\dfrac{-8}{27}\\ =>\left(3x-1\right)^3=\dfrac{\left(-2\right)^3}{3^3}\\ =>\left(3x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\\ =>3x-1=-\dfrac{2}{3}\\ =>3x=-\dfrac{2}{3}+1\\ =>3x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:3\\ =>x=\dfrac{1}{9}\)

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

\(\Rightarrow\) \(3x-1=\dfrac{-2}{3}\)

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

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

\(x=\dfrac{1}{3}\div3\)

\(x=\dfrac{1}{9}\)

Vậy \(x=\dfrac{1}{9}\)

\(a,A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\)

Để A nguyên thì: 3 ⋮ x - 2

=> x - 2 ∈ Ư(3) = {1; -1; 3; -3} 

=> x ∈ {3; 1; 5; -1} 

\(b,B=\dfrac{2x-1}{x+5}=\dfrac{\left(2x+10\right)-11}{x+5}=\dfrac{2\left(x+5\right)-11}{x+5}=2-\dfrac{11}{x+5}\)

Để B nguyên thì 11 ⋮ x + 5

=> x + 5 ∈ Ư(11) = {1; -1; 11; -11}

=> x ∈ {-4; -6; 6; -16} 

\(c,C=\dfrac{10x-9}{2x-3}=\dfrac{\left(10x-15\right)+6}{2x-3}=\dfrac{5\left(2x-3\right)+6}{2x-3}=5+\dfrac{6}{2x-3}\)

Để C nguyên thì 6 ⋮ 2x - 3

=> 2x - 3 ∈ Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}

Mà: 2x - 3 luôn lẻ

=> 2x - 3 ∈ {1; -1; 3; -3}

=> 2x ∈ {4; 2; 6; 0}

=> x ∈ {2; 1; 3; 0} 

Đặt \(P=x^3+y^3+z^3-3xyz\)

\(=x^3+\left(y+z\right)^3-3yz\left(y+z\right)-3xyz\)

\(=\left(x+y+z\right)\left[x^2-x\left(y+z\right)+\left(y+z\right)^2\right]-3yz\left(x+y+z\right)\)

\(=3\left(x^2+y^2+z^2+2yz-xy-xz\right)-9yz\)

\(=3\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=3\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)\right]\)

\(=3\left[9-3\left(xy+yz+zx\right)\right]\)

Do \(0\le x,y,z\le2\Rightarrow\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge0\)

\(\Rightarrow xyz+\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge0\)

\(\Rightarrow2\left(xy+yz+zx\right)-4\left(x+y+z\right)+8\ge0\)

\(\Rightarrow2\left(xy+yz+zx\right)\ge4.3-8=4\)

\(\Rightarrow xy+yz+zx\ge2\)

\(\Rightarrow P\le3.\left[9-3.2\right]=9\)

\(P_{max}=9\) khi \(\left(x;y;z\right)=\left(0;1;2\right)\) và các hoán vị của chúng

Với điều kiện đề bài \(0< x,y,z< 2\) thì biểu thức ko tồn tại GTLN

Biểu thức chỉ tồn tại GTLN khi \(0\le x,y,z\le2\) (có dấu = ở biên)

\(\Leftrightarrow\left(4x^2y^2+4xy+1\right)+\left(x^2-6x+9\right)=1\)

\(\Leftrightarrow\left(2xy+1\right)^2+\left(x-3\right)^2=1\)

Do \(2xy+1\) luôn lẻ với mọi x;y nguyên \(\Rightarrow2xy+1\ne0\Rightarrow\left(2xy+1\right)^2\ge1;\forall x;y\)

\(\Rightarrow\left(2xy+1\right)^2+\left(x-3\right)^2\ge1;\forall x;y\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(2xy+1\right)^2=1\\\left(x-3\right)^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

\(\dfrac{5}{7\times12}+\dfrac{4}{12\times16}+\dfrac{3}{16\times19}+\dfrac{2}{19\times21}+\dfrac{1}{21\times22}\\ =\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{22}\\ =\dfrac{1}{7}-\dfrac{1}{22}\\ =\dfrac{15}{154}\)

\(\dfrac{5}{7\times12}+\dfrac{4}{12\times16}+\dfrac{3}{16\times19}+\dfrac{2}{19\times21}+\dfrac{1}{21\times22}\)

\(=\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{22}\)

\(=\dfrac{1}{7}-\dfrac{1}{22}\)

\(=\dfrac{22}{154}-\dfrac{7}{154}\)

\(=\dfrac{15}{154}\)

Xác suất đúng của mỗi đáp án là: \(\dfrac{1}{4}\)

Xác suất cả 2 câu đều đúng là: \(\dfrac{1}{4}.\dfrac{1}{4}=\dfrac{1}{16}\)

Giúp mình với ạ cầu xin mọi người luôn đó 

\(65180=60000+5000+100+80=6.10^4+5.10^3+1.10^2+8.10^1\)

\(101010=100000+1000+10=1.10^5+1.10^3+1.10^1\)

\(\overline{ab0cd}=\overline{a0000}+\overline{b000}+\overline{c0}+d=a.10^4+b.10^3+c.10^1+d.10^0\)