3+2x-1 =24 - [42-(22-1)]
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\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(\Leftrightarrow2^{x-1}=24-16+3-3\)
\(\Leftrightarrow x-1=3\)
hay x=4
\(a,\Rightarrow2^{x-1}=24-\left(16-3\right)-3\\ \Rightarrow2^{x-1}=24-13-3\\ \Rightarrow2^{x-1}=8=2^3\\ \Rightarrow x-1=3\Rightarrow x=4\\ b,\Rightarrow\left(19x+50\right):14=25-16=9\\ \Rightarrow19x+50=126\\ \Rightarrow x=4\)
3 + 2x - 1= 24 - [42 - (22 - 1)
3 + 2x - 1= 24 - [42 - 21]
3 + 2x - 1= 24 - 21
3 + 2x - 1= 3
3 + 2x = 3 + 1
3 + 2x = 4
2x = 4 - 3
2x =1
x = 1:2
x = 0,5
Vậy x = 0,5
3 + 2x - 1= 24 - [42 - (22 - 1)
3 + 2x - 1= 24 - [42 - 21]
3 + 2x - 1= 24 - 21
3 + 2x - 1= 3
3 + 2x = 3 + 1
3 + 2x = 4
2x = 4 - 3
2x =1
x = 1:2
x = 0,5
suy ra x = 0,5
\(42-\left(2x+32\right)-12:\left(-2\right)=6\)
\(42-2x-32+12:2=6\)
\(\left(42-32+6-6\right)-2x=0\)
\(10-2x=0\)
\(x=5\)
\(\left(x-15\right):\left(-5\right)-\left(-22\right)=24\)
\(\left(x-15\right):\left(-5\right)+22=24\)
\(\left(x-15\right):\left(-5\right)=2\)
\(x-15=2\cdot\left(-5\right)\)
\(x-5=-10\)
\(x=-5\)
42−(2x+32)−12:(−2)=6
42−2x−32+12:2=6
10−2x=0
(x−15):(−5)−(−22)=24
(x−15):(−5)+22=24
(x−15):(−5)=2
x−15=2·(−5)
x−5=−10
x=−5
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
41+42+43+44-21-22-23-24
=( 41- 21 )+ (42-22)+(43-23)+(44-24)
=20 + 20 +20 +20
=20 . 4
=80
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\) (sửa đề)
\(\Rightarrow3+2^{x-1}=24-\left[16-\left(4-1\right)\right]\)
\(\Rightarrow3+2^{x-1}=24-\left(16-3\right)\)
\(\Rightarrow3+2^{x-1}=24-13\)
\(\Rightarrow3+2^{x-1}=11\)
\(\Rightarrow2^{x-1}=11-3\)
\(\Rightarrow2^{x-1}=8\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1=4\)
Vậy \(x=4.\)
#\(Toru\)
3+2x-1=24-[16-(4-1)]
3+2x+1=24-(16-3)
3+2x+1=24-13
3+2x+1=11
2x+1=11-3
2x+1=8
2x+1=23
=> x+1=3
x=2
3+2\(^{x-1}\)=24-[16-3]
3+2\(^{x-1}\)=24-13=11
2\(^{x-1}\)=11-3=8=2\(^3\)
=>x-1=3
=>x=4