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

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

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

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

\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

___________

\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

Bạn ghi lại đa thức đi bạn

a, 2^x-1 = 64

=> 2^x-1 = 2⁶

=> x - 1 = 6

=> x = 6 + 1 

=> x = 7

b, 3^2x-10 = 81

=> 3^2x-10 = 3⁴

=> 2x - 10 = 4

=> 2x = 14

=> x = 7

c, 2^2x+3 = 1024

=> 2^2x+3 = 2¹⁰

=> 2x + 3 = 10

=> 2x = 7

=> x = \(\frac{7}{2}\)

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

\(\left(2x-1\right)^6=64^2\)

\(\Leftrightarrow\left(2x-1\right)^6=4^6\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=4\\2x-1=-4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{3}{2}\end{array}\right.\)

\(\left(2x-1\right)^6=64^2\)

\(\Rightarrow\left(2x-1\right)^6=4^6\)

\(\Rightarrow2x-1=\pm4\)

+) \(2x-1=4\Rightarrow x=\frac{5}{2}\)

+) \(2x-1=-4\Rightarrow x=\frac{-3}{2}\)

Vậy \(x=\frac{5}{2}\) hoặc \(x=\frac{-3}{2}\)

(2x-1)⁶ = 64²

<=> (2x-1)⁶ = 4⁶

<=> 2x -1 = 4

<=> 2x = 5

<=> x = \(\frac{5}{2}\)

\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)