Ta có:

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

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

\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{x^3-y^3}\)

\(=\dfrac{2x^2+2y^2-4xy}{x^3-y^3}\)

\(=\dfrac{2x^2-2xy-2xy+2y^2}{x^3-y^3}\)

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

\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{2x-2y}{x^2+xy+y^2}\)

\(\dfrac{2}{x}\) + \(\dfrac{1}{y}\) = \(\dfrac{1}{6}\) (\(x;y\) \(\in\) N*)

\(\dfrac{2}{x}\)         = \(\dfrac{1}{6}\) - \(\dfrac{1}{y}\)

\(\dfrac{2}{x}\)         = \(\dfrac{y-6}{6y}\)

\(x\)          = 2: \(\dfrac{y-6}{6y}\)

\(x\)         = \(\dfrac{12y}{y-6}\)

Vì \(x\); y \(\in\) N* nên 12\(y\) ⋮ y - 6 ( và y > 6)

12y ⋮ y - 6 ⇔ 12y - 72 + 72 ⋮ y - 6 ⇔ 12.(y-6) + 72 ⋮ y - 6 ⇔ 72⋮ y - 6  72 = 2³.3² 

Ư(72) = { 1; 2; 3; 4; 6; 8; 9; 12; 18; 24; 36; 72}

Lập bảng ta có:

Theo bảng trên ta có các cặp số tự nhên \(x\); y thỏa mãn đề bài lần lượt là:

(\(x\);y) =(84;7); (48;8); (36;9); (30;10);(34;12); (21;14); (20;15);(18;18);

(16;24); (15; 30); (14;42);(13;78)

\(\dfrac{2}{x}+\dfrac{1}{y}=\dfrac{1}{6}\left(x;y\inℕ^∗\right)\)

\(\Leftrightarrow\dfrac{2y+x}{xy}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(2y+x\right)=xy\)

\(\Leftrightarrow12y+6x=xy\)

\(\Leftrightarrow12y-xy+6x=0\)

\(\Leftrightarrow y\left(12-x\right)+6x-72+72=0\)

\(\Leftrightarrow-y\left(x-12\right)+6\left(x-12\right)=-72\)

\(\Leftrightarrow\left(x-12\right)\left(6-y\right)=-72\)

\(\Leftrightarrow\left(x-12\right);\left(6-y\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-8;8;-9;9;-18;18;-24;24;-36;36;-72;72\right\}\)

Lập bảng sẽ ra \(\left(x;y\inℕ^∗\right)\) cần tìm...

Đặt 1/x=a; 1/y=b

Hệ phương trình trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{2}{3}\\\dfrac{1}{4}a+\dfrac{1}{3}b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=2\\15a+20b=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15b+15b=30\\15b+20b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5b=18\\a+b=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{18}{5}\\a=\dfrac{64}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{18}\\y=\dfrac{15}{64}\end{matrix}\right.\)

x\(\le\)11

=>3(x-1)-2(x-2)<=6/4(x-3)

=>3x-3-2x+4<=3/2x-9/2

=>-1/2x<=-9/2-1=-11/2

=>x>=11

a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)

\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)

\(\Rightarrow64x+96=72x-24\)

\(\Rightarrow8x=120\Rightarrow x=15\)

b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)

\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)

\(\Rightarrow221x-34=152x+380\)

\(\Rightarrow69x=414\Rightarrow x=6\)

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

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

\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1