a) x²³= 64 . x²⁰

x²³: x²⁰= 64

x³= 64

=> x³= 4³

=> x =4

Vậy...

a)\(x^{23}=64.x^{20}\)

\(\Leftrightarrow\frac{x^{23}}{x^{20}}=64\)

\(\Leftrightarrow x^3=64\Rightarrow x=4\)

b)\(\left(4x-3\right)^4=3-4x\)

\(\Leftrightarrow\left(3-4x\right)^4=3-4x\)

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

\(\Leftrightarrow3-4x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

<=>   \(\sqrt{64\left(x+1\right)}-\sqrt{25\left(x+1\right)}+\sqrt{4\left(x+1\right)}=20\)

<=> \(8\sqrt{\left(x+1\right)}-5\sqrt{\left(x+1\right)}+2\sqrt{\left(x+1\right)}=20\)

<=>   . \(5\sqrt{\left(x+1\right)}=20\)

<=>  \(\sqrt{\left(x+1\right)}=4\)

=> x+1=16

=> x=15

a) \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b) \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c) \(9-6x+x^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

a: Ta có: \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b: Ta có: \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

hay \(x=\dfrac{1}{2}\)

c: ta có: \(x^2-6x+9=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

hay x=3

a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50

b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1

c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10

d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13

ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => 4x² - 12x + 9 - x² - 10x - 25 = 0 

=> 3x² - 22x - 16 = 0

=> (x - 8)(3x + 2) = 0

=> x - 8 = 0 => x = 8

hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3

Vậy x = 8 ; x = -2/3

c/ => x³ - x² - 4x² + 8x - 4 = 0 

=> x³ - 5x² + 8x - 4 = 0 

=> (x - 2)² (x - 1) = 0

=> (x - 2)² = 0 => x - 2 = 0 => x = 2

hoặc x - 1 = 0 => x = 1 

Vậy x = 2 ; x = 1

đầu tiên ta tính nhẩm

2565.65-565,64

=2565.(64+1)-565.64

=(2565.64+2565)-565.64

=2565.64+2565-565.64

=64.(2565-565)+2565

=64.2000+2565

=128000+2565

=130565

áp dụng

10^4x+1+30565=2565.65-565.64

10^4x+1+30565=130565

10^4x+1=130565-30565=100000

10^4x+1=10^5

suy ra 4x+1=5

          4x=5-1=4

          4x=4 vậy x=1

tick cho mik nha mỏi tay quá đi mất

393 NHÉ

a) \(x^3+64=0\)

\(x^3=0-64\)

\(x^3=-64\)

\(x^3=-4^3\)

\(\Rightarrow x=-4\)

b)Tương tự

\(\left(x+1\right)^2=81\)

\(\Rightarrow\left(x+1\right)^2=9^2\)

\(\Rightarrow x+1=9\)

\(\Rightarrow x=9-1=8\)

Vậy x = 8

b, \(\left(x+5\right)^3=-64\)

\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=\left(-4\right)-5\)

\(\Rightarrow x=-9\)

Vậy x = -9

c, \(\left(2x-3\right)^2=9\)

\(\Rightarrow\left(2x-3\right)^2=3^2\)

\(\Rightarrow2x-3=3\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

Vậy x = 3

d, \(\left(4x+1\right)^3=27\)

\(\Rightarrow\left(4x+1\right)^3=3^3\)

\(\Rightarrow4x+1=3\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)