Bài 1 : Viết các đa thức sau dưới dạng lập phương của một tổng hoặc lập phương của một hiệu

a,8x3+12x2y+6xy2+y38x3+12x2y+6xy2+y3

= (2x)³ + 3.(2x)².y + 3.2x.y² + y³

= ( 2x + y )³
b,x3+3x2+3x+1x3+3x2+3x+1

= x³ + 3.x².1 + 3.x.1² + 1³

=(x + 1)³

c, x3−3x2+2x−1x3−3x2+2x−1

= x³ - 3.x².1+ 3.x.1² - 1³

= (x - 1)³

d,27+27y2+9y4+y6

= 3³ + 3.3².y² + 3.3.y4 + (y²)³

= ( 3 + y² ) ³

1111x99

\(\left(3x+1\right)^3=64\\ \Leftrightarrow\left(3x+1\right)^3=\left(4\right)^3\\ \Leftrightarrow3x+1=4\\ 3x=4-1\\ 3x=3\\ \Rightarrow x=3:3\\ \Rightarrow x=1\)

\(\left(3x+1\right)^3=64\)

mà 4³ = 64

=> 3x + 1 = 4

3x = 4 - 1

3x = 3

x = 3 : 3 = 1

vậy x = 1

Bài dễ vậy mà bạn không biết làm sao ==

x=1

\(m,x^3+48x=12x^2+64\)

\(x^3+48x-12x^2-64=0\)

\(\left(x-4\right)^3=0\)

\(x=4\)

\(n,x^3-3x^2+3x=1\)

\(x^3-3x^2+3x-1=0\)

\(\left(x-1\right)^3=0\)

\(x=1\)

\(\Leftrightarrow x^3+48x-12x^2-64=0\)0

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)-12x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(x-4\right)^3=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Trả lời:

m, x³ + 48x = 12x² + 64

<=> x³ + 48x - 12x² - 64  = 0

<=> x³ - 3.x².4 + 3.x.4² - 4³ = 0

<=> ( x - 4 )³ = 0

<=> x - 4 = 0

<=> x = 4

Vậy x = 4 là nghiệm của pt.

n, x³ - 3x² + 3x = 1

<=> x³ - 3x² + 3x - 1 = 0

<=> ( x - 1 )³ = 0

<=> x - 1 = 0

<=> x = 1

Vậy x = 1 là nghiệm của pt.

Ta có: \(\left(-\dfrac{3}{4}\right)^{3x-1}=-\dfrac{27}{64}\)

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

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

Vậy: \(x=\dfrac{4}{3}\)

  \(\left(-\dfrac{3}{4}\right)^{3x-1}=-\dfrac{27}{64}\)

  \(\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^3\)

\(\Rightarrow3x-1=3\)

     \(3x=3+1\)

     \(3x=4\)

\(\Rightarrow x=\dfrac{4}{3}\)

\(b,\left(5x-1\right)^2:2=8\\ \Leftrightarrow\left(5x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\end{matrix}\right.\\ c,\left(1-3x\right)^3=-64\\ \Leftrightarrow1-3x=-4\\ \Leftrightarrow3x=5\\ \Leftrightarrow x=\dfrac{5}{3}\)

a, (2x-3)³ = -64

=> (2x-3)³ = -4³

=> 2x-3=-4

=> 2x = -1

=> x = -1 : 2

=> x = -1/2

b, (2x-3)² =25

=>  (2x-3)² =5^2

=> 2x-3 = 5

=> 2x = 8

=> x = 4

c, (3x-4)² =36

=> (3x-4)² =6²

=> 3x-4 = 6

=> 3x = 10

=> x = 3.(3)

d, 2^x+1 = 64

=> 2^x+1  = 2⁶

=> x+1 = 6

=> x = 5

a/ (2x - 3)³ = -64 => (2x - 3)³ = (-4)³ =>  2x - 3 = -4 => 2x = -1 => x = -1/2

b/ (2x - 3)² = 25 => (2x - 3)² = 5² => 2x - 3 = 5 => 2x = 8 => x = 4

c/ (3x - 4)² = 36 => (3x - 4)² = 6² => 3x - 4 = 6 => 3x = 10 => x = 10/3

d/ 2^x+1 = 64 => 2^x+1 = 2⁶ => x + 1 = 6 => x = 5

\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)

\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)

a,

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)

\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

b,

( 3x + 1)^3 = 64

(3x + 1)^3 = 4^3

(3x + 1) = 4

3x = 4 - 1

3x = 3

x = 3 : 3

x = 1

c,

( 2x - 3)^4 = 81

( 2x - 3) ^4 = 3^4

(2x - 3) = 3

2x = 3 + 3

2x = 6

x = 6: 2

x = 3