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23 tháng 10 2015

(23:4).2x+1=64

(8:4).2x+1=64

2.2x+1=64

2x+1=64:2

2x+1=32

2x+1=25

=>x+1=5

x=5-1

x=4

31 tháng 10 2017

(2^3.4).2^x+1=64

(8:4).2^x+1=64

2.2^x+1=64

4^x+1=64

4^x+1=4^3

x+1=3

x=3-1          =)x=2

8 tháng 7 2016

a, (2x-3)3 = -64

=> (2x-3)3 = -43

=> 2x-3=-4

=> 2x = -1

=> x = -1 : 2

=> x = -1/2

b, (2x-3)=25

=>  (2x-3)=5^2

=> 2x-3 = 5

=> 2x = 8

=> x = 4

c, (3x-4)=36

=> (3x-4)=62

=> 3x-4 = 6

=> 3x = 10

=> x = 3.(3)

d, 2x+1 = 64

=> 2x+1  = 26

=> x+1 = 6

=> x = 5

8 tháng 7 2016

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

1: \(=x^4-8x^3+24x^2-32x+16+x^4-12x^3+54x^2-108x+81-1\)

\(=2x^4-20x^3+78x^2-140x+96\)

\(=2\left(x-3\right)\left(x-2\right)\left(x^2-5x+8\right)\)

2: \(=x^4-4x^3+6x^2-4x+1+x^4+12x^3+54x^2+108x+81-512\)

\(=2x^4+8x^3+60x^2+104x-430\)

\(=2\left(x^4+4x^3+30x^2+52x-215\right)\)

1 tháng 8 2016

Ta có: \(4^{2x+4}=4^3\)

\(\Rightarrow\)\(2x+4=3\)

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

1 tháng 8 2016

Ta có: \(3^{x-1}+3^{x-2}=108\)

\(\Rightarrow\)\(3^{x-1}+3^{x-2}=2^2.3^3\)

\(\Rightarrow\)\(3^{x-2}=3^3\)

\(\Rightarrow\)\(x-2=3\)

\(\Rightarrow x=5\)

16 tháng 12 2018

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

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

\(x^4+2x^2-24=\left(x^2+6\right)\cdot\left(x^2-4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+6\right)\)

30 tháng 8 2021

đẳng cấp

 

a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)

\(\Leftrightarrow2x=\dfrac{1}{64}\)

hay \(x=\dfrac{1}{128}\)

8 tháng 8 2017

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

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

3 tháng 12 2017

\(5-\left|3x-1\right|=3\)

\(\left|3x-1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

                 vậy \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

\(\left|x+\frac{3}{4}\right|-5=-2\)

\(\left|x+\frac{3}{4}\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{15}{4}\end{cases}}\)

\(\left(1-2x\right)^2=9\)

\(\left(1-2x\right)^2=3^2\)

\(\Rightarrow1-2x=3\)

\(\Rightarrow2x=-2\)

\(\Rightarrow x=-1\)

vậy \(x=-1\)

\(\left(x+5\right)^3=-64\)

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

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

\(\Rightarrow x=-9\)

vậy \(x=-9\)

\(\left(2x+1\right)^2=\frac{4}{9}\)

\(\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\)

\(\Rightarrow2x+1=\frac{2}{3}\)

\(\Rightarrow2x=\frac{-1}{3}\)

\(\Rightarrow x=\frac{-1}{6}\)

vậy \(x=-\frac{1}{6}\)