\(4^{x+2}=244+3.4^{x-1}\)

\(\Rightarrow4^{x-1}.4^3=244+3.4^{x-1}\)

\(\Rightarrow4^{x-1}\left(4^3-3\right)=244\)

\(\Rightarrow4^{x-1}.61=244\)

\(\Rightarrow4^{x-1}=4\Rightarrow x-1=1\Rightarrow x=2\)

Chúc bạn học tốt.

Ta có : (x - 1)² = (x - 1)⁴

=> (x - 1)⁴ - (x - 1)² = 0

=> (x - 1)².[(x - 1)² - 1] = 0

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

Nếu x - 1 = 0 => x = 1

Nếu x - 1 = 1 => x = 2

Nếu x - 1 = - 1 => x = 0

Vậy \(x\in\left\{0;1;2\right\}\) 

b) 5 ^{- 1} . 25^x = 125

=> \(\frac{1}{5}.25^x=125\)

=> 25^x = 625

=> 25^x = 25²

=> x = 2

Vậy x = 2

a) \(\left(x-1\right)^2=\left(x-1\right)^4\Leftrightarrow1=\left(x-1\right)^2\)\(\Leftrightarrow x-1=1\Leftrightarrow x=2\)

b) \(5^{-1}.25^x=125\Leftrightarrow5.25^{x-1}=125\Leftrightarrow25^{x-1}=25\)\(\Rightarrow x-1=1\Leftrightarrow x=2\)

chinh lại câu c.

x+2/5=2-3x

x+3x=2-2/5

4x=8/5

x=2/5

a. /2x-1/=/x+2/

2x-1= x+2 hoặc 2x-1=-x-2

x=3 hoặc 3x=-1

x=3 hoặc x=-1/3

b. 2.\(3^x\).\(3^2\)=18

\(3^x\).18=18

\(3^x\)=1

\(3^x\)=\(3^0\)

x=0

c.x+2/5=2-3x

x+3x=2-2/5

4x=3/5

x=3/20

mới học lớp 6 à

Ta có: (x - 2,5)²⁰¹⁴ + |x + y + 0,5| = 0

Mà: (x - 2,5)²⁰¹⁴ lớn hơn hoặc bằng 0 và |x + y + 0,5| cũng lớn hơn hoặc bằng 0

Nên để thỏa mãn đẳng thức đã cho thì: (x - 2,5)²⁰¹⁴ = 0 và |x + y + 0,5| = 0 => x - 2,5 = 0 và x + y + 0,5 = 0

Với x - 2,5 = 0 => x = 2,5

Thay x = 2,5 vào x + y + 0,5 = 0 => y = -3

• Khoảng 1: x<2013x is less than 2013𝑥<2013
• • |x−2013|=−(x−2013)=2013−xthe absolute value of x minus 2013 end-absolute-value equals negative open paren x minus 2013 close paren equals 2013 minus x|𝑥−2013|=−(𝑥−2013)=2013−𝑥
  • |x−2014|=−(x−2014)=2014−xthe absolute value of x minus 2014 end-absolute-value equals negative open paren x minus 2014 close paren equals 2014 minus x|𝑥−2014|=−(𝑥−2014)=2014−𝑥
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(2013−x)+(2014−x)+(2015−x)=6042−3xcap B equals open paren 2013 minus x close paren plus open paren 2014 minus x close paren plus open paren 2015 minus x close paren equals 6042 minus 3 x𝐵=(2013−𝑥)+(2014−𝑥)+(2015−𝑥)=6042−3𝑥 (Giảm dần)
• Khoảng 2: 2013≤x<20142013 is less than or equal to x is less than 20142013≤𝑥<2014
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=−(x−2014)=2014−xthe absolute value of x minus 2014 end-absolute-value equals negative open paren x minus 2014 close paren equals 2014 minus x|𝑥−2014|=−(𝑥−2014)=2014−𝑥
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(x−2013)+(2014−x)+(2015−x)=2016−xcap B equals open paren x minus 2013 close paren plus open paren 2014 minus x close paren plus open paren 2015 minus x close paren equals 2016 minus x𝐵=(𝑥−2013)+(2014−𝑥)+(2015−𝑥)=2016−𝑥 (Giảm dần)
• Khoảng 3: 2014≤x<20152014 is less than or equal to x is less than 20152014≤𝑥<2015
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=x−2014the absolute value of x minus 2014 end-absolute-value equals x minus 2014|𝑥−2014|=𝑥−2014
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(x−2013)+(x−2014)+(2015−x)=x−2012cap B equals open paren x minus 2013 close paren plus open paren x minus 2014 close paren plus open paren 2015 minus x close paren equals x minus 2012𝐵=(𝑥−2013)+(𝑥−2014)+(2015−𝑥)=𝑥−2012 (Tăng dần)
• Khoảng 4: x≥2015x is greater than or equal to 2015𝑥≥2015
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=x−2014the absolute value of x minus 2014 end-absolute-value equals x minus 2014|𝑥−2014|=𝑥−2014
  • |x−2015|=x−2015the absolute value of x minus 2015 end-absolute-value equals x minus 2015|𝑥−2015|=𝑥−2015
  • B=(x−2013)+(x−2014)+(x−2015)=3x−6042cap B equals open paren x minus 2013 close paren plus open paren x minus 2014 close paren plus open paren x minus 2015 close paren equals 3 x minus 6042𝐵=(𝑥−2013)+(𝑥−2014)+(𝑥−2015)=3𝑥−6042 (Tăng dần) 

1. Tìm giá trị nhỏ nhất:
2. • Giá trị B giảm đến x=2014x equals 2014𝑥=2014 (B = 2016 - 2014 = 2) rồi bắt đầu tăng.
   • Giá trị nhỏ nhất của B là 2, đạt được khi xx𝑥 nằm trong khoảng [2014,2015]open bracket 2014 comma 2015 close bracket[2014,2015].