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\(a)C_6H_6 + 3Cl_2 \to C_6H_6Cl_6\\ n_{Cl_2}= \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{C_6H_6} = \dfrac{1}{3}n_{Cl_2} = \dfrac{0,1}{3}(mol)\\ m_{C_6H_6} = \dfrac{0,1}{3}.78 = 26(gam)\\ b) n_{C_6H_6Cl_6} = n_{C_6H_6} = \dfrac{0,1}{3}(mol)\\ m_{C_6H_6Cl_6} = \dfrac{0,1}{3}.291 = 97(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 127 - 2(-2x + 4)² = -1
2(-2x + 4)² = 127 + 1
2(-2x + 4)² = 128
(-2x + 4)² = 128 : 2
(-2x + 4)² = 64
-2x + 4 = 8 hoặc -2x + 4 = -8
*) -2x + 4 = 8
-2x = 8 - 4
-2x = 4
x = 4 : (-2)
x = -2
*) -2x + 4 = -8
-2x = -8 - 4
-2x = -12
x = -12 : (-2)
x = 6
b) -6² - 3.(-4 - x)³ = -39
3(-4 - x)³ = -6² + 39
3(-4 - x)³ = -36 + 39
3(-4 - x)³ = 3
(-4 - x)³ = 3 : 3
(-4 - x)³ = 1
(-4 - x)³ = 1³
-4 - x = 1
x = -4 - 1
x = -5
c) (-7x - 35).(2x² - 32) = 0
-7x - 35 = 0 hoặc 2x² - 32 = 0
*) -7x - 35 = 0
-7x = 0 + 35
-7x = 35
x = 35 : (-7)
x = -5
*) 2x² - 32 = 0
2x² = 0 + 32
2x² = 32
x² = 32 : 2
x² = 16
x = -4; x = 4
Vậy x = -5; x = -4; x = 4
d) (2018 + 2x)(4x² + 8) = 0
2018 + 2x = 0 (do 4x² + 8 > 0)
2x = 0 - 2018
2x = -2018
x = -2018 : 2
x = -1014
e) -145 - 4(3 - 2x)² = -245
4(3 - 2x) = -145 + 245
4(3 - 2x) = 100
3 - 2x = 100 : 4
3 - 2x = 25
2x = 3 - 25
2x = -22
x = -22 : 2
x = -11
g) 4(-x + 1)³ - (-5)² = -525
4(-x + 1)³ - 25 = -525
4(-x + 1)³ = -525 + 25
4(-x + 1)³ = -500
(-x + 1)³ = -500 : 4
(-x + 1)³ = -125
(-x + 1)³ = (-5)³
-x + 1 = -5
-x = -5 - 1
-x = -6
x = 6
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e: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)\(\)
\(\Leftrightarrow x^2+3x=x\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=3(4x503+1).2(4x503+2)
A=3(4x503)+3.2(4x503)+22
A=(.....1)+3.(....6)+4
A=(....4).(.....0)
A=(.....0)
k cho mình được ko
Ta có A = 3^2013.2^2014
A= 3^2012+1 . 2^2012+2
A= 3^2012.3.2^2012.2^2
A= (3^4)^503 .( 2^4)^503 .(3.4)
A= (...1)^503.(..6)^503.(..2)
A= (..1).(..6).(..2)
A=..2