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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
a) 2x + 34 = 72
2x=38
x=19
b) (4+x) . 2 phần 3 =9
4+x=\(\frac{27}{2}\)
x=\(\frac{19}{2}\)
c) (2x + 1 ) 3 =36
2x+1=12
2x=11
x=\(\frac{11}{2}\)
d ) (x + 5 ) : 4 =12
x+5=48
x=43
e ) 1 phần 2 x -- 4 = 15
\(\frac{1}{2}x\)=19
x=38
a) 42 - x = 5
x = 42 - 5
x = 37
b) 15.(x + 4) = 75
x + 4 = 75 : 15
x + 4 = 5
x = 5 - 4
x = 1
d) x + 18 = 12
x = 12 - 18
x = -6
e) 2x - 15 = -19
2x = (-19) + 15
2x = -4
x = -4 : 2
x = -2
f) 36 : (x^3 - 12) = -3
x^3 - 12 = 36 : (-3)
x^3 - 12 = -12
x^3 = 0
=> x = 0
g)x + 15 = 35
x = 35 - 15
x = 20
h) 2x - 13 = 3^2
2x - 13 = 9
2x = 9 + 13
2x = 22
x = 22 : 2
x = 11
1, => [ 462 + 642 - 2x ] : 2 = 531
1104 -2x = 1062
2x = 42
x =21
Bài 1:
a) Ta có: \(-5x+32=\left(-2\right)^3\)
\(\Leftrightarrow-5x+32=-8\)
\(\Leftrightarrow-5x=-40\)
hay x=8
Vậy: x=8
b) Ta có: \(-2x+36=6\)
\(\Leftrightarrow-2x=6-36=-30\)
hay x=15
Vậy: x=15
e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
b,-2x+36=6
tương đương -2x=-30
tương đương x=15
a, -5x+32=(-2)^3
tương đương -5x+32=8
tương đương -5x=-24
tương đương x=24/5
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
\(a.\left[\left(8x-12\right):4\right]\cdot3^3=3^6\\ \left(8x-12\right):4=3^6:3^3\\ \left(8x-12\right):4=27\\ 8x-12=27\cdot4\\ 8x-12=108\\ 8x=108+12\\ 8x=120\\ x=15\\ b.41-2^{x+1}=9\\ 2^{x+1}=32\\ 2^{x+1}=2^5\\=> x+1=5\\ =>x=4\)
a,[(8x-12):4].33=36
(8x-12):4=36:33
(8x-12):4=33
(8x-12):4=9
8x-12=9.4
8x-12=36
8x=36+12
8x=48
x=6
Vậy x=6
b,41-2x+1=9
2x+1=41-9
2x+1=32
2x+1=25
x+1=5
x=5-1
x=4
Vậy x=4
Chúc bạn học tốt:33
\(\left[\left(2x-4\right)+9\right]^3=3^6\)
\(\Leftrightarrow2x-4=0\)
hay x=2