Tìm x
a) 2x:25=1 b)5x+x=39-311:39
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a) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
b) \(2^x:2^5=16\)
\(\Leftrightarrow2^x:2^5=2^4\)
\(\Leftrightarrow2^x=2^4.2^5\)
\(\Leftrightarrow2^x=2^9\)
\(\Leftrightarrow x=9\)
c) \(7x-x=5^{21}:5^{19}+3.2^2-7^0\)
\(\Leftrightarrow6x=5^2+3.4-1\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
d) \(7x-2x=6^{17}:6^{15}+44:11\)
\(\Leftrightarrow5x=6^2+4\)
\(\Leftrightarrow5x=40\)
\(\Leftrightarrow x=8\)
a)⇔6x=39-32
⇔6x=30
⇔ x=5
b)2x:25=16
⇔2x=24.25
⇔ 2x=29
⇔ x=9
c)⇔6x=52+3.22-1
⇔ 6x= 36
⇔ x=6
d)⇔5x=62+4
⇔ 5x=40
⇔ x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3x=81
nên x=27
b: \(5\cdot4^x=80\)
\(\Leftrightarrow4^x=16\)
hay x=2
c: \(2^x=4^5:4^3\)
\(\Leftrightarrow2^x=2^4\)
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow6x=30\)
hay x=5
b: \(\Leftrightarrow6x=25+12-1=36\)
hay x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow8x=108+12=120\)
hay x=15
b: \(\Leftrightarrow6x=60\)
hay x=10
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b: Ta có: 7x-8=713
nên 7x=721
hay x=103
c: Ta có: x-36:18=12
nên x-2=12
hay x=14
d: Ta có: (x-36):18=12
nên x-36=216
hay x=252
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^3\cdot4^{x-1}=64\)
\(\Leftrightarrow4^{x-1}=1\)
\(\Leftrightarrow x-1=0\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,5x-16=40-1\)
<=>\(5x=40-1+16\)
<=>\(5x=55\)
<=>\(x=11\)
\(b,x-10=-25\)
<=>\(x=\left(-25\right)-10\)
<=>\(x=-35\)
\(c,-12+x=-30\)
<=>\(x=\left(-30\right)+12\)
<=>\(x=-18\)
a) 5x-16=40-1
5x-16=39
5x=39+16
5x=55
x=55:5
x=11
b)x-10=(-25)
x=(-25)+10
x=(-15)
c) -12+x= -30
x= -30-(-12)
x= -30+12
x= -18
d) 2x +12 = -40+ 6
2x+12=(-34)
2x=(-34)-12
2x=(-46)
x=(-46):2
x=(-23)
e)125:(3x-13)=25
3x-13=125:25
3x-13=5
3x=5+13
3x=18
x=18:3
x=6
f)541+(218-x)=735
218-x=735-541
218-x=194
x=218-194
x= 24
g) 3(2x+1)-19=14
3(2x+1)=14+19
3(2x+1)=33
2x+1=33:3
2x+1=11
2x=11-1
2x=10
x=10:2
x=5
h)175-5(x+3)=85
5(x+3)=175-85
5(x+3)=90
x+3=90:5
x+3=18
x=18-3
x=15
i)8x+(-3)=39
8x=39-(-3)
8x=42
x=42:8
x=42/8
L)2x+4x=36+(-6)
6x=30
x=30:6
x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
b,
\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)
c,
\(a)2^x:2^5=1\)
\(2^x:32=1\)
\(2^x=1\cdot32\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(b)5x+x=39-\frac{3^{11}}{3^9}\)
\(6x=39-3^{11-9}\)
\(6x=39-3^2\)
\(6x=39-9\)
\(6x=30\)
\(x=\frac{30}{6}\)
\(x=5\)
ý a x=5
ý b x=5