2/5 - x/7 = 0,25 + 2/-9
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Lời giải:
a. $(x.0,25+1999).2000=(53+1999).2000$
$x.0,25.2000+1999.2000=53.2000+1999.2000$
$x.0,25.2000=53.2000$
$x.0,25=53$
$x=53:0,25=212$
b.
$(5457+x:2):7=1075$
$5457+x:2=1075\times 7=7525$
$x:2=7525-5457=2068$
$x=2068\times 2=4136$
c.
$1-(\frac{12}{5}+x-\frac{8}{9}): \frac{16}{9}=0$
$(\frac{12}{5}+x-\frac{8}{9}):\frac{16}{9}=1$
$\frac{12}{5}+x-\frac{8}{9}=1.\frac{16}{9}=\frac{16}{9}$
$\frac{68}{45}+x=\frac{16}{9}$
$x=\frac{16}{9}-\frac{68}{45}=\frac{4}{15}$

\(\begin{array}{l}a)x + 0,25 = \frac{1}{2}\\x = \frac{1}{2} - 0,25\\x = \frac{1}{2} - \frac{1}{4}\\x = \frac{2}{4} - \frac{1}{4}\\x = \frac{1}{4}\end{array}\)
Vậy \(x = \frac{1}{4}\)
\(\begin{array}{l}b)x - \left( { - \frac{5}{7}} \right) = \frac{9}{{14}}\\x = \frac{9}{{14}} + \left( { - \frac{5}{7}} \right)\\x = \frac{9}{{14}} + \left( { - \frac{{10}}{{14}}} \right)\\x = \frac{{ - 1}}{{14}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{14}}\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

1. \(x-8+32=68\)
\(x-8=68-32\)
\(x-8=36\)
\(x=36+8\)
\(x=44\)
_____
2. \(x+8+32=68\)
\(x+8=68-32\)
\(x+8=36\)
\(x=36-8\)
\(x=28\)
_____
3. \(98-x+34=43\)
\(98-x=43-34\)
\(98-x=9\)
\(x=98-9\)
\(x=89\)
_____
4. \(98+x-34=43\)
\(98+x=43+34\)
\(98+x=77\)
\(x=77-98\)
\(x=-21\)
_____
5. \(x:5:4=800\)
\(x:5=800.4\)
\(x:5=3200\)
\(x=3200.5\)
\(x=16000\)
_____
6. \(18+x=384:8\)
\(18+x=48\)
\(x=48-18\)
\(x=30\)
_____
7. \(\left(84,6-2\times x\right):3,02=5,1\)
\(84,6-2.x=5,1.3,02\)
\(84,6-2.x=15,402\)
\(2.x=84,6-15,402\)
\(2.x=69,198\)
\(x=69,198:2\)
\(x=34,599\)
_____
8. \(x\times5=120:6\)
\(x.5=20\)
\(x=20:5\)
\(x=4\)
_____
9. \(\left(15\times24-x\right):0,25=100:0,25\)
\(\left(15.24-x\right):0,25=400\)
\(15.24-x=400.0,25\)
\(15.24-x=100\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
1,x-8+32=68
=>x-8=68-32=36
=>x=36+8=42
2,x+8+32=68
=>x+8=68-32=36
=>x=36-8=28
3,98-x+34=43
=>98-x=43-34=9
=>x=98-9=89
4,98+x-34=43
=>98+x=43+34=77
=>x=77-98=-21
5,x:5:4=800
=>x:20=800
=>x=800x20=16000
6,18+x=384:8=48
=>x=48-18=30
7,(84,6-2x):3,02=5,1
=>84,6-2x=5,1x3,02=15,402
=>2x=84,6-15,402=69,198
=>x=69,198:2=34,599
8,5x=120:6=20
=>x=20:5=4
9,(15x24-x):0,25=100:0,25=400
=>360-x=400x0,25=100
=>x=360-100=260.


1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
\(\frac25-\frac{x}{7}\) = 0,25 + \(\frac{2}{-9}\)
\(\frac15-\frac{x}{7}\) = \(\frac14\) - \(\frac29\)
\(\frac15-\frac{x}{7}\) = \(\frac{9}{36}-\) \(\frac{8}{36}\)
\(\frac15-\frac{x}{7}\) = \(\frac{1}{36}\)
\(\frac{x}{7}\) = \(\frac25-\frac{1}{36}\)
\(\frac{x}{7}\) = \(\frac{72}{180}\) - \(\frac{5}{180}\)
\(\frac{x}{7}\) = \(\frac{67}{180}\)
\(x=\frac{67}{180}\times7\)
\(x\) = \(\frac{469}{180}\)
Vậy \(x=\frac{469}{180}\)
Ta có: \(\dfrac{2}{5}-\dfrac{x}{7}=0,25+\dfrac{2}{-9}\)
=>\(\dfrac{2}{5}-\dfrac{x}{7}=\dfrac{1}{4}-\dfrac{2}{9}=\dfrac{9}{36}-\dfrac{8}{36}=\dfrac{1}{36}\)
=>\(\dfrac{x}{7}=\dfrac{2}{5}-\dfrac{1}{36}=\dfrac{72}{180}-\dfrac{5}{180}=\dfrac{67}{180}\)
=>\(x=\dfrac{67}{180}\cdot7=\dfrac{469}{180}\)