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\(a)\frac{1}{7}x-\frac{1}{2}x+\frac{5}{7}x=-\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{7}-\frac{1}{2}+\frac{5}{7}\right)x=-\frac{1}{2}\)
\(\Rightarrow\left(\frac{2}{14}-\frac{7}{14}+\frac{10}{14}\right)x=-\frac{1}{2}\)
\(\Rightarrow\frac{5}{14}x=-\frac{1}{2}\)
\(\Rightarrow x=-\frac{1}{2}:\frac{5}{14}\)
\(\Rightarrow x=-\frac{1}{2}.\frac{14}{5}\)
\(\Rightarrow x=-\frac{7}{5}\)
\(b)(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{49.51})x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{49}-\frac{1}{51}\right)x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{51}\right)x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{51}{561}-\frac{11}{561}\right)x=-\frac{1}{3}\)
\(\Rightarrow\frac{40}{561}x=-\frac{1}{3}\)
\(\Rightarrow x=-\frac{1}{3}:\frac{40}{561}\)
\(\Rightarrow x=-\frac{1}{3}.\frac{561}{40}\)
\(\Rightarrow x=-\frac{187}{40}\)
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\(a,\frac{23}{15}-\frac{2}{3}x=0,2\)
\(\frac{2}{3}x=\frac{4}{3}\)
\(x=2\)
\(b,\frac{x}{3}=\frac{2}{3}+\left(\frac{-1}{7}\right)\)
\(\Rightarrow\frac{x}{3}=\frac{11}{21}\)
\(\Rightarrow x.21=3.11\)
\(\Rightarrow21x=33\)
\(\Rightarrow x=\frac{11}{7}\)
a. \(\frac{23}{15}-\frac{2}{3}x=0,2\)
\(\frac{23}{15}-\frac{2}{3}x=\frac{1}{5}\)
\(\frac{2}{3}x=\frac{23}{15}-\frac{1}{5}\)
\(\frac{2}{3}x=\frac{4}{3}\)
\(x=\frac{4}{3}:\frac{2}{3}\)
\(x=2\)
Vậy \(x=2\)
b. \(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\frac{x}{3}=\frac{11}{21}\)
\(\frac{7x}{21}=\frac{11}{21}\)
\(\Rightarrow7x=11\)
\(x=11:7\)
\(x=\frac{11}{7}\)
Vậy\(x=\frac{11}{7}\)
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
1: =>x=3/5-1/5=2/5
b: =>x/3=5/8+1/8=3/4
=>x=9/4
3: =>10/3x=3+1/4+6+3/4=10
=>x=10:10/3=3
a, -19 - x = -20
x = -19 - (-20)
x = -19 + 20
x = 1
b, 5x - 6 = 3x + 12
5x - 6 - 3x = 12
5x - 3x = 12 + 6
(5 - 3)x = 18
2x = 18
x = 18 : 2
x = 9
c, 15 - 3 (x - 1) = 8 - 2x
15 - 3 (x - 1) + 2x = 8
-3x - 3 - 2x = 8 - 15
-3x - 3 - 2x = -7
-3x - 2x - 3 = 7
-3x - 2x = 7 + 3
(-3 - 2) x = 10
-5x = 10
x = 10 : (-5)
x = -2
d, (5x - 6)2 = 16
(5x - 6)2 = 42
=> 5x - 6 = 4
5x = 4 + 6
5x = 10
x = 10 : 5
x = 2
f, 26 - | x + 9 | = 13
| x + 9 | = 26 - 13
=> | x + 9 | = 13
=> x + 9 = +- 13
* Với x + 9 = 13
x = 13 - 9
x = 4
* Với x + 9 = -13
x = -13 - 9
x = -22
Vậy x = {4;-22}
e, | 3 + x | = 19
=> 3 + x = +- 19
* Với 3 + x = 19
x = 19 - 3
x = 16
* Với 3 + x = -19
x = -19 - 3
x = -22
Vậy x = {16;-22}
a, X = -19+20=1
b, (5-3)X = 18
2X = 18
=> X = 9
c, 3X + 3 -2X = 7
X+3 =7
X = 4
f, |X+9| = 13
ta có 2 trường hợp:
TH1: X+9 = 13
=> X= 4
TH2 : X+9 = -13
=> X= -22
e, ta có 2 trường hợp:
TH1: 3+X = 19
=> X= 16
TH2: 3+X = -19
=> X= -22
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a) Ta có: \(x:3\dfrac{1}{15}=1\dfrac{1}{2}\)
\(\Leftrightarrow x:\dfrac{46}{15}=\dfrac{3}{2}\)
nên \(x=\dfrac{3}{2}\cdot\dfrac{46}{15}=\dfrac{23}{5}\)
b) Ta có: \(\left(1\dfrac{1}{5}-0.2\right)\cdot x=1\)
\(\Leftrightarrow\left(1+\dfrac{1}{5}-\dfrac{1}{5}\right)\cdot x=1\)
hay x=1
a, \(x:3\dfrac{1}{15}=1\dfrac{1}{12}\Leftrightarrow x:\dfrac{46}{15}=\dfrac{13}{12}\Leftrightarrow x=\dfrac{299}{90}\)
b, \(\left(1\dfrac{1}{5}-0,2\right)x=1\Leftrightarrow\left(\dfrac{6}{5}-\dfrac{1}{5}\right)x=1\Leftrightarrow x=1\)