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1)
a) 2x + 5 = 3⁴ : 3²
2x + 5 = 3²
2x + 5 = 9
2x = 9 - 5
2x = 4
x = 4 : 2
x = 2
b) (3x - 24).73 = 2.74
(3x - 24).73 = 148
3x - 24 = 148/73
3x = 148/73 + 24
3x = 1900/73
x = 1900/73 : 3
x = 1900/219
c) [3.(42 - x)] + 15 = 23.3
126 - 3x + 15 = 69
141 - 3x = 69
3x = 141 - 69
3x = 72
x = 72 : 3
x = 24
d) 126 + (132 - x) = 300
132 - x = 300 - 126
132 - x = 174
x = 132 - 174
x = -42
2)
a) 120 - (x + 55) = 60
x + 55 = 120 - 60
x + 155 = 60
x = 60 - 55
x = 5
b) (7x - 11).3 = 25.52 + 200
(7x - 11).3 = 1500
7x - 11 = 1500 : 3
7x - 11 = 500
7x = 500 + 11
7x = 511
x = 511 : 7
x = 73
c) 2x + 2x + 4 = 544
4x = 544 - 4
4x = 540
x = 540 : 4
x = 135
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
f) (x – 5)2 = 9
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
(7x-11)3=25.52+200
=> (7x-11)3=800+200
=> (7x-11)3=1000
=> (7x-11)3=103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
(2x-15)5=(2x-15)3
=> (2x-15)5 - (2x-15)3 = 0
=> (2x-15)3 . [ (2x-15)2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}2x=15\\2x=16\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
Mà x thuộc N
=> x = 8
(3x-5)10=(3x-5)9
=> (3x-5)10 - (3x-5)9 = 0
=> (3x-5)9 .[ (3x-5) - 1 ] = 0
=> \(\orbr{\begin{cases}\left(3x-5\right)^9=0\\\left(3x-5\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-5=0\\3x-5=1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=5\\3x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
Mà x thuộc N
=> x = 2
a)(7x-11)^3=1000
(7x-11)^3=10^3
7x-11 =10
7x =10+11=21
x =21:7=3
a, \(3^x.4=324\)
\(\Rightarrow\) \(3^x=81\)
\(\Rightarrow\) \(3^x=3^4\)
\(\Rightarrow\) \(x=4\)
b, \(x^{100}=x\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
c, \(\left(3x-1\right)^3=125\)
\(\Rightarrow\) \(\left(3x-1\right)^3=5^3\)
\(\Rightarrow\) \(3x-1=5\)
\(\Rightarrow\) \(3x=6\)
\(\Rightarrow\) \(x=2\)
d, \(2x-15=17\)
\(\Rightarrow\) \(2x=32\)
\(\Rightarrow\)\(x=16\)