Bài 1:
a.
$=(\frac{-2}{17}+\frac{-15}{17})+(\frac{15}{19}+\frac{4}{19})+\frac{13}{20}$
$=\frac{-17}{17}+\frac{19}{19}+\frac{13}{20}=-1+1+\frac{13}{20}=\frac{13}{20}$
b.
$=\frac{7}{19}.\frac{8}{11}+\frac{7}{19}.\frac{3}{11}+\frac{12}{19}$
$=\frac{7}{19}(\frac{8}{11}+\frac{3}{11})+\frac{12}{19}$
$=\frac{7}{19}.1+\frac{12}{19}=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1$
c.
$=2,07+3,005-12,005+4,23$
$=(2,07+4,23)-(12,005-3,005)$
$=6,3-9=-2,7$
d.
$=\frac{4}{25}+\frac{5}{4}-\frac{1}{4}=\frac{4}{25}+\frac{4}{4}=\frac{4}{25}+1=\frac{29}{25}$
Bài 1:
a: \(\dfrac{-2}{17}+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{13}{20}+\dfrac{4}{19}\)
\(=\left(-\dfrac{2}{17}-\dfrac{15}{17}\right)+\left(\dfrac{15}{19}+\dfrac{4}{19}\right)+\dfrac{13}{20}\)
\(=-1+1+\dfrac{13}{20}=\dfrac{13}{20}\)
b: \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}:\dfrac{11}{3}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{19}{19}=1\)
c: \(\left(2,07+3,005\right)-\left(12,005-4,23\right)\)
\(=2,07+3,005-12,005+4,23\)
=(2,07+4,23)-9
=6,3-9
=-2,7
d: \(\left(-\dfrac{2}{5}\right)^2+\dfrac{1}{2}\left(4,5-2\right)-25\%\)
\(=\dfrac{4}{25}+\dfrac{1}{2}\cdot2,5-0,25\)
=0,16-0,25+1,25
=1,25-0,09
=1,16
Bài 2:
a: \(\dfrac{1}{2}x-\dfrac{3}{4}=-1\dfrac{1}{4}\)
=>\(\dfrac{1}{2}x=-\dfrac{5}{4}+\dfrac{3}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}\)
=>x=-1
b: \(\left(\dfrac{1}{2}+2x\right)\left(2x-3\right)=0\)
=>\(\left[{}\begin{matrix}2x+\dfrac{1}{2}=0\\2x-3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
c: \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)
=>\(x-\dfrac{2}{3}-\dfrac{3}{2}=\dfrac{1}{4}\)
=>\(x-\dfrac{13}{6}=\dfrac{1}{4}\)
=>\(x=\dfrac{1}{4}+\dfrac{13}{6}=\dfrac{3}{12}+\dfrac{26}{12}=\dfrac{29}{12}\)
d: \(\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1=-\dfrac{7}{3}\)
=>\(x\cdot\dfrac{7}{6}=-\dfrac{7}{3}\)
=>\(x=-\dfrac{7}{3}:\dfrac{7}{6}=-\dfrac{7}{3}\cdot\dfrac{6}{7}=-2\)
e: \(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)
=>\(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{2}\\2x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=0\\2x=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
f: \(\dfrac{-2}{5}< \dfrac{x}{15}< \dfrac{1}{6}\)
=>\(\dfrac{-12}{30}< \dfrac{2x}{30}< \dfrac{5}{30}\)
=>-12<2x<5
=>-6<x<2,5
mà x nguyên
nên \(x\in\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)
Bài 2:
a.
$\frac{1}{2}x=-1\frac{1}{4}+\frac{3}{4}=\frac{-1}{2}$
$x=\frac{-1}{2}:\frac{1}{2}=-1$
b.
$(\frac{1}{2}+2x)(2x-3)=0$
$\Rightarrow \frac{1}{2}+2x=0$ hoặc $2x-3=0$
$\Rightarrow x=\frac{-1}{4}$ hoặc $x=\frac{3}{2}$
c.
$2(\frac{1}{2}x-\frac{1}{3})=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}$
$\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}: 2= \frac{7}{8}$
$\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frsc{29}{24}$
$x=\frac{29}{24}: \frac{1}{2}=\frac{29}{12}$
Bài 2:
d.
$\frac{1}{2}x+\frac{2}{3}x-1=-3\frac{1}{3}$
$x(\frac{1}{2}+\frac{2}{3})=-3\frac{1}{3}+1$
$x.\frac{7}{6}=\frac{-7}{3}$
$x=\frac{-7}{3}: \frac{7}{6}=-2$
e.
$\frac{1}{4}-(2x+\frac{1}{2})^2=0$
$(2x+\frac{1}{2})^2=\frac{1}{4}=(\frac{1}{2})^2=(\frac{-1}{2})^2$
$\Rightarrow 2x+\frac{1}{2}=\frac{1}{2}$ hoặc $2x+\frac{1}{2}=\frac{-1}{2}$
$\Rightarrow 2x=0$ hoặc $2x=-1$
$\Rightarrow x=0$ hoặc $x=\frac{-1}{2}$
f.
$\frac{-2}{5}< \frac{x}{15}< \frac{1}{6}$
$\frac{-12}{30}< \frac{2x}{30}< \frac{5}{30}$
$\Rightarrow -12< 2x< 5$
$\Rightarrow -6< x< 2,5$
$\Rightarrow x\in \left\{-5; -4; -3; -2; -1; 0; 1; 2\right\}$