Bài 1: Mở rộng khái niệm phân số

\(a,ĐKXĐ:n-1\ne0\Leftrightarrow x\ne1\)

b, \(n=0\Rightarrow A=\dfrac{3}{0-1}=-3\)

\(n=5\Rightarrow A=\dfrac{3}{5-1}=\dfrac{3}{4}\)

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

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)

\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)

Vô lí => không có x,y thỏa mãn

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)

nên \(\dfrac{x}{y}=\dfrac{3}{7}\)

b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)

hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

Cảm ơn mn

B12

a) ta có \(\dfrac{170}{17}=10\Rightarrow x=9\)

b) ta có  \(\dfrac{13}{3}=4,333...\Rightarrow x=4\)

B13

a) x>\(\dfrac{-12}{13}\Rightarrow x=-1\)

b) ta có: \(\dfrac{-45}{15}\Rightarrow x>-3\Rightarrow x=-2\)

Bài 12: 

a. \(x< -\dfrac{170}{17}=-10\rightarrow\) x nguyên max = -11

b. \(x< -\dfrac{13}{3}=-4,\left(3\right)\rightarrow\)x nguyên max là -5

Bài 13: 

a. \(x>-\dfrac{12}{13}=-0,926..\rightarrow\)x nguyên min = 0 

b. \(x>-\dfrac{45}{15}=-3\rightarrow\)x nguyên min = -2

\(x^2-9x=0\\ \Rightarrow x\left(x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

Vậy \(x\in\left\{0;9\right\}\)

x2− 9x=0

⇒x (x − 9) = 0

⇒[x = 0x − 9 = 0

⇒[ x = 0x = 9

Vậy 

   \(\dfrac{1}{10}-\left(\dfrac{1}{3}\right)^2:\dfrac{5}{9}\)

= \(\dfrac{1}{10}-\dfrac{1}{9}:\dfrac{5}{9}\)

= \(\dfrac{1}{10}-\dfrac{1}{9}.\dfrac{9}{5}\)

= \(\dfrac{1}{10}-\dfrac{1}{5}\)

= \(\dfrac{1}{10}-\dfrac{2}{10}\)

= \(-\dfrac{1}{10}\)

\(\dfrac{1}{10}-\left(\dfrac{1}{3}\right)^2:\dfrac{5}{9}\)

=\(\dfrac{1}{10}-\dfrac{1}{9}:\dfrac{5}{9}\)

=\(\dfrac{1}{10}-\dfrac{1}{9}.\dfrac{9}{5}\)

=\(\dfrac{1}{10}-\dfrac{1}{5}\)

=\(\dfrac{1}{10}-\dfrac{2}{10}\)

=\(\dfrac{-1}{10}\)=-0.1

Ta có: \(\dfrac{1}{10}-\left(\dfrac{1}{3}\right)^2:\dfrac{5}{9}\)

\(=\dfrac{1}{10}-\dfrac{1}{9}:\dfrac{5}{9}\)

\(=\dfrac{1}{10}-\dfrac{1}{9}\cdot\dfrac{9}{5}\)

\(=\dfrac{1}{10}-\dfrac{1}{5}\)

\(=\dfrac{1}{10}-\dfrac{2}{10}\)

\(=-\dfrac{1}{10}\)

\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}=\dfrac{7}{9}\)

Ta có: \(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)

\(=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{2}{3}-\dfrac{5}{3}\right)+\dfrac{7}{9}\)

\(=1-1+\dfrac{7}{9}=\dfrac{7}{9}\)

\(\dfrac{k}{-5}\)

đáp án là \(\dfrac{k}{-5}\)

Chúc bạn học tốt

\(\dfrac{k}{-5}\)