48-48-0+12+41
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![](https://rs.olm.vn/images/avt/0.png?1311)
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a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
22,48 x 48 + 22,48 : 0,1 + 22,48 x 41 + 22,48
= 22,48 x 48 + 22,48 x 10 + 22,48 x 41 + 22,48
= 22,48 x ( 48 + 10 + 41 + 1 )
= 22,48 x 100
= 2248
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =-48+27+56-48-27-36
=-96+20
=-76
b: =23-57+57-33=-10
c: =-98+12-159-12-41
=-98-200
=-298
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{8\times12\times32+4\times27\times24+96\times41}{16\times48-48\times9+4\times3\times12}\)
A = \(\dfrac{96\times32+96\times27+96\times41}{16\times48-48\times9+48\times3}\)
A = \(\dfrac{96\times\left(32+27+41\right)}{48\times\left(16-9+3\right)}\)
A = \(\dfrac{96\times100}{48\times10}\)
A = \(\dfrac{20}{1}\) \(\times\) \(\dfrac{48\times10}{48\times10}\)
A = 20
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, $-28\cdot41+100\cdot141+41\cdot(-72)$
$=41\cdot[(-28)+(-72)]+100\cdot141$
$=41\cdot(-100)+(-100)\cdot(-141)$
$=(-100)\cdot[41+(-141)]$
$=(-100)\cdot(-100)$
$=100\cdot100$
$=10000$
b, $35\cdot(65-6)-65\cdot(35+6)$
$=35\cdot65+35\cdot(-6)-65\cdot35-65\cdot6$
$=(35\cdot65-65\cdot35)+[35\cdot(-6)+65\cdot(-6)]$
$=0+(-6)\cdot(35+65)$
$=-6\cdot100$
$=-600$
c, $-48+48\cdot(-78)-21\cdot48$
$=-48\cdot1+(-48)\cdot78+(-48)\cdot21$
$=-48\cdot(1+78+21)$
$=-48\cdot(79+21)$
$=-48\cdot100$
$=-4800$
![](https://rs.olm.vn/images/avt/0.png?1311)
b. 1500(x-7)=0
x-7=0
x=7
c. (2x-4)(48-12x)=0
2x-4=0 hoặc 48-12x=0
x=2 hoặc x=4
d. (x+12)(x-1)=0
x+12=0 hoặc x-1=0
x=-12 hoặc x=1
bài 2 :
a . 128-3(x+4)=23
3(x+4)=105
x+4=35
x=31
b. [(14X+26).3+55]:5=35
(14x+26).3+55=175
(14x+26).3=120
14x+26=40
14x=14
x=1
d. 720:[41-(2X-5)]=23.5
41-(2x-5)=720:(23.5)
41-(2x-5)=144/23
2x-5=799/23
2x=914/23
x=457/23
b, 1500.(x – 7) = 0
<=>1500x-10500=0
<=>1500x=10500
<=>x=7
Vậy x=7
c,(2.x – 4).(48 – 12.x) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy x=2 hoặc x=4
d, (x + 12).(x – 1) =0
\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)
Vậy x=-12 hoặc x=1
Bài 2:
a) 128- 3(x+ 4) = 23
\(\Leftrightarrow\)128-(3x+12)=23
\(\Leftrightarrow\)128-3x-12=23
\(\Leftrightarrow\)116-3x=23
\(\Leftrightarrow\)3x=116-23
\(\Leftrightarrow\)3x=93
\(\Leftrightarrow\)x=31
Vậy x=31
b) [(14x+ 26). 3+ 55]: 5= 35
\(\Leftrightarrow\)(14x+ 26). 3+ 55=175
\(\Leftrightarrow\)42x+78+55=175
\(\Leftrightarrow\)42x+133=175
\(\Leftrightarrow\)42x=175-133
\(\Leftrightarrow\)42x=42
\(\Leftrightarrow\)x=1
Vậy x=1
d, 720: [41- (2x- 5)]= 23. 5
\(\Leftrightarrow\)720: 41- (2x- 5)=115
\(\Leftrightarrow\)41-(2x- 5)=720:115
\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)
\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)
\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)
\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)
Vậy x=\(\dfrac{457}{23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
48-48-0+12+41=53
48 - 48 - 0 + 12 + 41
= 0 - 0 + 12 + 41
= 53
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