a) 

\(\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\left(x\ne1;x\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{2}{y}=-2\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{5}{x-1}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{1}{y}=-1\\x-1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-1-1=-2\\x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=2\end{matrix}\right.\left(tm\right)\)

b) 

\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\left(x\ne2;y\ne-1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2}+\dfrac{2}{y+1}=6\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y+1}=5\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=1\\\dfrac{2}{x-2}+1=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\\dfrac{2}{x-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x-2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\left(tm\right)\)

c)

\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\left(x\ne2;y\ne1\right) \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=4\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{5}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{3}{5}=2\\y-1=\dfrac{5}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=2-\dfrac{3}{5}=\dfrac{7}{5}\\y=\dfrac{5}{3}+1=\dfrac{8}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+2=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\left(tm\right)\)

a) Hiệu số phần bằng nhau là:

7 - 4 = 3 (phần)

Đáy lớn là:

CD = 150 : 3 x 7 = 350 (m)

Đáy bé là:

AB = 150 : 3 x 4 = 200 (m) 

Diện tích mảnh đất là: 

\(80\cdot\dfrac{350+200}{2}=22000\left(m^2\right)\)

b) \(EC=\dfrac{1}{5}CD=\dfrac{1}{5}\cdot350=70\left(m\right)\) 

Diện tích phần nuôi cá là:

\(80\cdot\dfrac{200+70}{2}=10800\left(m^2\right)\)

c) Diện tích ruộng mà bình nhận được là:

\(22000-10800=11200\left(m^2\right)\)

Tìm n để thoả mãn điều gì thế em???

a) 

\(\left\{{}\begin{matrix}\dfrac{1}{3x}+\dfrac{1}{3y}=\dfrac{1}{4}\\\dfrac{5}{6x}+\dfrac{1}{y}=\dfrac{2}{3}\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{5}{6x}+\dfrac{1}{y}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{5}{6x}=\dfrac{3}{4}-\dfrac{2}{3}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{6x}=\dfrac{1}{12}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x=12\Leftrightarrow x=2\\\dfrac{1}{2}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\\dfrac{1}{y}=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\left(tm\right)\)

b) 

\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=2\\2\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=16\end{matrix}\right.\left(x,y\ne0\right) \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=2\\\dfrac{1}{x}+\dfrac{1}{y}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}=10\\\dfrac{1}{x}+\dfrac{1}{y}=8\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\5+\dfrac{1}{y}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\\dfrac{1}{y}=8-5=3\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{1}{3}\end{matrix}\right.\left(tm\right)\)

\(\overline{ab4}-\overline{ab}=\overline{a48}\\\overline{ab0}+4-\overline{ab}=\overline{a00}+48\\ \overline{ab}\times10-\overline{ab}=\overline{a00}+48-4\\ \overline{ab}\times9=\overline{a44} \) (*)

Để tìm được \(\overline{ab}\) là STN có 2 chữ số, hiển nhiên \(\overline{a44}\) phải chia hết cho 9

Suy ra: a + 4 + 4 cũng phải chia hết cho 9

hay a + 8 chia hết cho 9

Mà a là số có 1 chữ số nên a = 1

Thay vào biểu thức (*) :

\(\overline{1b}\times9=144\\ \overline{1b}=144:9\\ \overline{1b}=16\\ b=6\) (nhận)

Vậy: a=1 và b=6 hay \(\overline{ab}=16\)

Hoàn thiện phép tính:

\(164-16=148\)

\(\dfrac{11}{8}\left[\left(\dfrac{-5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\left(\dfrac{-5}{11}\cdot\dfrac{8}{13}+\dfrac{-5}{11}\cdot\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\dfrac{-5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left(\dfrac{-5}{11}+\dfrac{-2}{11}\right)+\dfrac{3}{4}\\ =\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}\\ =\dfrac{-7}{8}+\dfrac{6}{8}\\ =\dfrac{-1}{8}\)

Bài 3:

\(a.\dfrac{3x+2}{x-3}=\dfrac{3x-9+11}{x-3}=\dfrac{3\left(x-3\right)+11}{x-3}=3+\dfrac{11}{x-3}\)

Để bt nguyên thì 11 ⋮ x - 3

=> x - 3 ∈ Ư(11) = {1; -1; 11; -11} 

=> x ∈ {4; 2; 14; -8} 

\(b.\dfrac{x-2}{3x-2}\) nguyên \(\Rightarrow\dfrac{3\left(x-2\right)}{3x-2}=\dfrac{3x-6}{3x-2}=\dfrac{3x-2-4}{3x-2}=1-\dfrac{4}{3x-2}\)

Để bt nguyên thì: 4 ⋮ 3x - 2

=> 3x - 2 ∈ Ư(4) = {1; -1; 2; -2; 4; -4}

=> 3x ∈ {3; 1; 4; 0; 5; -2} 

=> x ∈ {1; `1/3`; `4/3`; 0;`5/3`;`-2/3`} 

Mà: x nguyên => x ∈ {1;0}

Bài 4:

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

=>\(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)

=>\(3A+A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3+3^{100}-3^{99}+...+3^2-3+1\)

=>\(4A=3^{101}+1\)

=>\(A=\dfrac{3^{101}+1}{4}\)

(35 x 2 - 35:5 x 10) x (1 + 3 + 5 + ... + 19)

= [35 x 2 - 35 x (10 : 5) ] x (1 + 3 + 5 + ... + 19) 

= (35 x 2 - 35 x 2) x (1 + 3 + 5 +... + 19)

= (70 - 70) x (1 + 3 + 5 + .... + 19) 

= 0 x (1 + 3 + 5 + ... + 19)

= 0

   (35 x 2 - 35: 5 x 10) x (1 + 3 + 5 + ... + 17 + 19)

= (70 - 7 x 10) x (1 + 3 + 5 + ... + 17 + 19)

= (70 - 70) x (1 + 3 + 5 + ... + 17 + 19)

= 0 x (1 + 3 + 5 + ... +17 + 19)

= 0 

a) 

\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{101}-\dfrac{3}{13}}{\dfrac{6}{101}-\dfrac{6}{13}+\dfrac{6}{7}-\dfrac{6}{11}+6}\\ =\dfrac{\left(23+1\right)\cdot47-23}{24+47\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}-\dfrac{1}{13}\right)}{6\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}-\dfrac{1}{13}\right)}\\ =\dfrac{23\cdot47+47-23}{23\cdot47+24}\cdot\dfrac{3}{6}\\ =\dfrac{23\cdot47+24}{23\cdot47+24}\cdot\dfrac{1}{2}\\ =1\cdot\dfrac{1}{2}=\dfrac{1}{2}\)

b) 

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{19\cdot21}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{3}{19\cdot21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\dfrac{20}{21}\\ =\dfrac{10}{21}\)

Dễ dàng so sánh được:

\(-\dfrac{5}{3}< -\dfrac{3}{3}=-1=-\dfrac{5}{5}< -\dfrac{4}{5}< 0< \dfrac{1}{3}< \dfrac{5}{3}< \dfrac{6}{3}=2=\dfrac{4}{2}< \dfrac{7}{2}\)

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