=>16x+9y=840 và 210/x-210/y=7/4

=>16x=840-9y và 30/x-30/y=1/4

=>x=-9/16y+52,5 và (30y-30x)=xy/4

=>xy=120y-120x

=>y(-9/16y+52,5)=120y-120(-9/16y+52,5)

=>-9/16y^2+52,5y-120y+120(-9/16y+52,5)=0

=>-9/16y^2-67,5y-67,5y+6300=0

=>y=40 hoặc y=-280

=>x=30 hoặc x=210

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

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 nên đăng vào mấy h mà nhìu ng onl í 

chứ h này ko còn ai đou ạ

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

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

a: =>4x^2-2x-x-4=0

=>4x^2-3x-4=0

=>\(x=\dfrac{3\pm\sqrt{73}}{8}\)

b: =>x^2+x-x+2=5x-4

=>x^2+2=5x-4

=>x^2-5x+6=0

=>x=2(loại) hoặc x=3(nhận)

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

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

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

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

\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow x=3\)

Sửa đề: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
Ta có: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)

\(\Leftrightarrow\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}+1=\dfrac{x+2001}{15}+1+\dfrac{x+2014}{2}+1\)

\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}=\dfrac{x+2016}{15}+\dfrac{x+2016}{2}\)

\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}-\dfrac{x+2016}{15}-\dfrac{x+2016}{2}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\ne0\)

nên x+2016=0

hay x=-2016

Vậy: S={-2016}

x= 3/20

4/30

ủa thế ai mới đúng

ĐKXĐ:\(x\ne-2\)

\(\dfrac{1}{x+2}-1=\dfrac{5x+7}{x+2}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{5x+7}{x+2}=1\\ \Leftrightarrow\dfrac{1-5x-7}{x+2}=1\\ \Leftrightarrow-5x-6=x+2\\ \Leftrightarrow x+2+5x+6=0\\ \Leftrightarrow6x+8=0\\ \Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)