khó à nghen

\(a,5x-16=40-1\)

<=>\(5x=40-1+16\)

<=>\(5x=55\)

<=>\(x=11\)

\(b,x-10=-25\)

<=>\(x=\left(-25\right)-10\)

<=>\(x=-35\)

\(c,-12+x=-30\)

<=>\(x=\left(-30\right)+12\)

<=>\(x=-18\)

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

a) x¹⁰ = 1^x

=> x¹⁰ = 1

=> x¹⁰ = 1¹⁰ => x = 1

b) x¹⁰ = x

=> x¹⁰ - x = 0

=> x(x⁹ - 1) = 0

=> \(\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

c) x⁵ = 32 => x⁵ = 2⁵ => x = 2

d) 3^x = 81 => 3^x = 3⁴ => x = 4

e) 25^x = 125²

=> 25^x = (5³)²

=> (5²)^x = 5⁶

=> 5^2x = 5⁶

=> 2x = 6 => x = 3

f) (2x - 15)⁵ = (2x - 15)³

=> (2x - 15)⁵ - (2x - 15)³ = 0

=> (2x - 15)³ [(2x - 15)² - 1] = 0

=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{cases}}\)

+) 2x - 15 = 1 => 2x = 16 => x = 8

+) 2x - 15 = -1 => 2x = 14 => x = 7

Vậy x = 8,x = 7

Nếu bài dễ thì cố gắng tự làm bạn nhé.

Bài 1:

a) \(2x+3=5\)

\(\Leftrightarrow2x=5-3=2\)

\(\Leftrightarrow x=2:2=1\)

Vậy: x=1

b) \(\left(2x-3\right)^2=9\)\(=3^2=\left(-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(3+3\right):2=3\\x=\left(-3+3\right):2=0\end{matrix}\right.\)

Vậy: \(x\in\left\{3;0\right\}\)

c) \(3x+4=2x-7\)

\(\Leftrightarrow2x-3x=7+4\)

\(\Leftrightarrow-1x=11\)

\(\Leftrightarrow x=-11\)

Vậy: x=-11

d) \(\left(4x-1\right)^3=27\)\(=3^3\)

\(\Rightarrow4x-1=3\)

\(\Leftrightarrow4x=3+1=4\)

\(\Leftrightarrow x=4:4=1\)

Vậy: x=1

e) \(\left(x-7\right).2=3.\left(2x+4\right)\)

\(\Leftrightarrow2x-14=6x+12\)

\(\Leftrightarrow2x-6x=12+14\)

\(\Leftrightarrow-4x=26\)

\(\Leftrightarrow x=\frac{26}{-4}=\frac{13}{-2}\)

Vậy: \(x=\frac{13}{-2}\)

a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

Ta có : (2x + 1)³ = 125

=> (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

Ta có : (4x - 1)² = 25 x 9

=> (4x - 1)² = 5².3²

=> (4x - 1)² = 15²

\(\Leftrightarrow\orbr{\begin{cases}4x-1=15\\4x-1=-15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=16\\4x=-14\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{7}{2}\end{cases}}\)

c) 2x - 49 = 5 . 3²

2x - 49 = 5 . 9

2x - 49 = 45

2x = 45 + 49

2x = 94

x = 94 : 2

x = 47