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`x+17=3^5:3^2`
`=>x+17=3^3`
`=>x+17=27`
`=>x=27-17`
`=>x=10`
__
`5.6^(x+1)-2.3^2=12`
`=>5.6^(x+1)-2.9=12`
`=>5.6^(x+1)-18=12`
`=>5.6^(x+1)=12+18=30`
`=>6^(x+1)=30:5`
`=>6^(x+1)=6`
`=>x+1=1`
`=>x=0`
A. \(\text{x + 17 = 3⁵ : 3²}\)
\(x+17=3^{5-2}\)
\(x+17=3^3\)
\(x+17=27\)
\(x=27-17\)
\(x=10\)
B.\(5\cdot6^{x+1}-2\cdot3^2=12\)
\(5\cdot6^{x+1}-2\cdot9=12\)
\(5\cdot6^{x+1}-18=12\)
\(5\cdot6^{x+1}=18+12\)
\(5\cdot6^{x+1}=30\)
\(6^{x+1}=\dfrac{30}{5}\)
\(6^{x+1}=6\)
\(x+1=1\)
\(x=0\)
\(x+1=1\) vì \(6=6^1\)
5/7-4/9=17/63
3/8+9/5=87/40
1-5/9=4/9
7/12-2/7+1/12=8/21
12/17-5/17-4/17=3/17
B=[2+(-7)]+[12+(-17)]+...+[102+(-107)]+112
B=-5+-5+...+-5+112
B=-5.503+112=-2403
`#3107.101107`
A.
`-7 + x = -3`
`=> x = -3 - (-7)`
`=> x = -3 + 7`
`=> x = 4`
Vậy, `x = 4`
B.
`-456 - x = 723`
`=> x = -456 - 723`
`=> x = -1179`
Vậy, `x = -1179`
C.
`-124-312 + x = -415`
`=> -436 + x = -415`
`=> x = -415 - (-436)`
`=> x = -415 + 436`
`=> x = 21`
Vậy, `x = 21`
D.
`-214+512-(-163)-x=-768- (-423)`
`=> 298 - (-163) - x = -345`
`=> 298 + 163 - x = -345`
`=> 461 - x = -345`
`=> x = 461 - (-345)`
`=> x = 461 + 345`
`=> x = 806`
Vậy, `x= 806.`
a: x-7=-3
=>x=-3+7=4
b: -456-x=723
=>x=-456-723=-1179
c: -124-312+x=-415
=>x-436=-415
=>x=21
d: -214+512-(-163)-x=-768-(-423)
=>461-x=-345
=>x=461+345=806
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Bài 31:
Gọi số mol KHCO3, CaCO3 là x, y (mol)
\(n_{BaCO_3}=\dfrac{1,97}{197}=0,01\left(mol\right)\)
PTHH: KHCO3 + HCl --> KCl + CO2 + H2O
x----------------------->x
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
y----------------------------->y
Ba(OH)2 + CO2 --> BaCO3 + H2O
0,01<---0,01
=> x + y = 0,01
a = 100x + 100y = 1 (g)
Bài 32:
Gọi số mol K2CO3, CaCO3 là a, b (mol)
=> 138a + 100b = 11,9 (1)
PTHH: K2CO3 + 2HCl --> 2KCl + CO2 + H2O
a--------------->2a------>a
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b-------------->b-------->b
=> \(a+b=\dfrac{2,24}{22,4}=0,1\) (2)
=> a = 0,05; b = 0,05
=> \(\left\{{}\begin{matrix}m_{KCl}=0,1.74,5=7,45\left(g\right)\\m_{CaCl_2}=0,05.111=5,55\left(g\right)\end{matrix}\right.\)
=> mmuối = 7,45 + 5,55 = 13 (g)