Bài 2: Cộng, trừ số hữu tỉ

ĐKXĐ: \(x\ne2\)

Ta có: \(\dfrac{-3}{x-2}=\dfrac{4}{5}\)

\(\Leftrightarrow x-2=\dfrac{-3\cdot5}{4}=-\dfrac{15}{4}\)

\(\Leftrightarrow x=-\dfrac{15}{4}+2=\dfrac{-15}{4}+\dfrac{8}{4}\)

hay \(x=-\dfrac{7}{4}\)

Vậy: \(x=-\dfrac{7}{4}\)

x=-7/4

\(x-\dfrac{1}{2}\left(1-x\right)\) > 0

\(\Leftrightarrow x-\dfrac{1}{2}+\dfrac{x}{2}>0\)

\(\Leftrightarrow\dfrac{2.x}{2}-\dfrac{1}{2}+x>0\)

\(\Leftrightarrow2x-1+x>0\)

\(\Leftrightarrow\) \(3x>1\)

\(\Leftrightarrow x>\dfrac{1}{3}\)

Vậy để \(x-\dfrac{1}{2}\left(1-x\right)>0\) thì \(x>\dfrac{1}{3}\)

\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{3}x-\dfrac{1}{4}\)

\(\dfrac{3}{2}x-\dfrac{1}{3}x=-\dfrac{1}{4}+\dfrac{2}{5}\)

\(\left(\dfrac{3}{2}-\dfrac{1}{3}\right)x=\dfrac{3}{20}\)

\(\dfrac{7}{6}x=\dfrac{3}{20}\)

\(x=\dfrac{9}{70}\)

\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{3}x-\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{1}{3}x=-\dfrac{1}{4}+\dfrac{2}{5}\)
\(x\left(\dfrac{3}{2}-\dfrac{1}{3}\right)=\dfrac{3}{20}\)
\(\dfrac{7}{6}x=\dfrac{3}{20}\)
\(x=\dfrac{3}{20}:\dfrac{7}{6}\)
\(x=\dfrac{9}{70}\)

\(\)Vậy \(x=\dfrac{9}{70}\)

\(\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)

\(=\dfrac{2}{145}.\left(\dfrac{1}{9}-\dfrac{13}{3}+1\right)\)

\(=\dfrac{2}{145}.\left(\dfrac{1}{9}-\dfrac{13}{3}+\dfrac{9}{9}\right)\)

\(=\dfrac{2}{145}.\left(-\dfrac{29}{9}\right)\)\(=-\dfrac{2}{45}\)

\(\left(5+\dfrac{1}{5}-\dfrac{2}{9}\right)-\left(2-\dfrac{1}{23}-\dfrac{73}{35}+\dfrac{5}{6}\right)-\left(8+\dfrac{2}{7}-\dfrac{1}{18}\right)\)

\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)

\(=\left(5-2-8\right)+\left(\dfrac{1}{5}+\dfrac{73}{35}-\dfrac{2}{7}\right)+\left(\dfrac{2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)

\(=\left(-5\right)+\left(\dfrac{7}{35}+\dfrac{73}{35}-\dfrac{10}{35}\right)+\left(\dfrac{4}{18}-\dfrac{15}{18}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)

\(=\left(-5\right)+2+\dfrac{-5}{9}+\dfrac{1}{23}\)

\(=\left(-3\right)+\dfrac{-5}{9}+\dfrac{1}{23}=\dfrac{-727}{207}\)

Vì (x+1).(x-2)<0 nên (x+1) và (x-2) khác dấu

TH1:

x+1<0→ x<-1

x-2>0→ x>2

⇒2<x<-1

⇒ko có x

TH2:

x+1>0→x>-1

x-2<0→x<2

⇒-1<x<2

⇒x∈{0;1

\(\left(x+1\right).\left(x-2\right)< 0\)

⇒ \(\left(x+1\right)\) và \(\left(x-2\right)\) khác dấu

mà \(\left(x+1\right)>\left(x-2\right)\)

⇒ \(\left(x+1\right)>0\) ; \(\left(x-2\right)< 0\)

\(\left(x+1\right)>0\) ⇒ \(x>-1\)

\(\left(x-2\right)< 0\) ⇒ \(x< 2\)

⇒ \(x\) ∈ \(\left\{0;1\right\}\)

\(x+x.y+y+1=2\)

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=2\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=2\)

vì x,y thuộc Z nên (x+1) và (y+1) \(\inƯ\left(2\right)=\left\{-1;1;2;-2\right\}\)

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

vậy \(S=\left\{\left(0;1\right),\left(1;0\right),\left(-2;-3\right),\left(-3;-2\right)\right\}\)

Ta có:

\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)

\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)

\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)

\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)

a)93/7:-3/7+178/7:3/7

3/7.(-93/7+178/7)=3/7.85/7=255/49

b)7.12⁸.5⁴-6⁹.20⁴/6¹⁰.10⁵-7.3⁹.80⁴

7.3⁸.2¹⁶.5⁴2⁹.3⁹.5⁴.2⁸:2¹⁰.3¹⁰.2⁵.5⁵-7.3⁹.2¹⁶.5⁴

=7².3³⁶.2⁴⁴.5¹⁷

\(=27\cdot\dfrac{-32}{243}\cdot\dfrac{625}{16}\cdot\dfrac{125}{2}=-\dfrac{78125}{9}\)