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a, \(\dfrac{3}{4}x=-\dfrac{9}{8}\)
x= \(-\dfrac{3}{2}\)
b, |x| + 0,25= 5,25
|x | = 5
=> x\(\in\){ +- 5}
Ko chắc đúng, kiểm tra trc khi làm
\(\dfrac{2x-1}{3}=\dfrac{-5}{0.6}\)
\(\Leftrightarrow2x-1=-25\)
hay x=-12
a, 11/13 - ( 5/42 - x ) = - (5/28 - 11/13)
11/13 - (5/42 - x) = - 5/28 + 11/13
- (5/42 - x) + 5/28 = -11/13 + 11/13
- 5/42 + x + 5/28 = 0
- 5/42 + x = 0 - 5/28
- 5/42 + x = - 5/28
x = -5/28 +5/42
x = - 5/84
b, / x + 4/15 \ - / - 3,75 \ = - / - 2,15 \
./ x + 4/15 \ - 3,75 = - 2,15
/ x + 4/15 \ = -2,15 + 3,75
/ x + 4/15 \ = 1,6
x + 4 / 15 = 1,6 hoặc x+ 4/15 = - 1,6
x = 1,6 - 4/15 x = - 1,6 -4/15
x = 4/3 x = -28/15
Vậy x = 4/3 hoặc x = - 28/15
c, ( 0,25 - 30% x ) . 1/3 = 1/4 - 31/6
( 1/4 - 3/10 x ) . 1/3 = - 59/12
( 1/4 - 3/10 x ) = - 59/12 : 1/3
1/4 - 3/10 x = - 59/4
3/10 x = 1/4 + 59/4
3/10 x = 15
x = 15 : 3/10
x = 50
d, ( x - 1/2 ) : 1/3 + 5/7 = 68/7
( x - 1/2 ) : 1/3 = 68/7 - 5/7
( x - 1/2 ) : 1/3 = 63/7
( x - 1/2 ) = 63/7 . 1/3
x -1/2 = 3
x = 3 + 1/2
x = 7/2
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
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) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
\(\begin{array}{l}a)5{x^3} + {x^3} = (5 + 1){x^3} = 6{x^3}\\b)\dfrac{7}{4}{x^5} - \dfrac{3}{4}{x^5} = \left( {\dfrac{7}{4} - \dfrac{3}{4}} \right){x^5} = \dfrac{4}{4}{x^5} = {x^5}\\c)( - 0,25{x^2}).(8{x^3}) = ( - 0,25.8).({x^2}.{x^3}) = - 2.{x^5}\end{array}\)
`# \text {nKaiz}`
`-4/5*x - (0,25 - x) = -13/3`
`=> -4/5x - 0,25 + x = -13/3`
`=> -4/5x + x = -13/3 + 0,25`
`=> 1/5x = -49/12`
`=> x = -49/12 \div 1/5`
`=> x = -245/12`
Vậy, `x = -245/12.`