\(125-2x=25\\ 2x=125-25\\ 2x=100\\ x=50\\ 100-\left(x+15\right)=27\\ x+15=100-27\\ x+15=73\\ x=73-15\\ x=58\)

\(125-2x=25\Leftrightarrow-2x=25-125\Leftrightarrow-2x=-100\Leftrightarrow x=\dfrac{-100}{-2}=50\)

\(100-\left(x+15\right)=27\Leftrightarrow-\left(x+15\right)=27-100\Leftrightarrow-\left(x+15\right)=-73\Leftrightarrow x+15=73\Leftrightarrow x=73-15\Leftrightarrow x=58\)

Sửa đề: \(\dfrac{16}{15}\rightarrow\dfrac{16}{25}\) 

Giải:

\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{16}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{4}{5}\right)^2\\\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{-4}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{4}{5}\\x-\dfrac{3}{5}=\dfrac{-4}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{-1}{5}\end{matrix}\right.\) 

\(\left(\dfrac{5}{3}-x\right)^3=\dfrac{1}{27}\) 

\(\Rightarrow\left(\dfrac{5}{3}-x\right)^3=\left(\dfrac{1}{3}\right)^3\) 

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

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

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

\(\left(4x-1\right)^3=27\Leftrightarrow4x-1=3\Leftrightarrow4x=4\Leftrightarrow x=1\)

\(\left(4x-1\right)^3=27\\ \Leftrightarrow\left(4x-1\right)^3=3^3\\ \Leftrightarrow4x-1=3\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

Ta có: \(\left(4x-1\right)^3=27\)

\(\Leftrightarrow4x-1=3\)

\(\Leftrightarrow4x=4\)

hay x=1

\(\left(2x+1\right)^3=-125\)

\(< =>\left(2x+1\right)^3=\left(-5\right)^3\)

\(< =>2x+1=-5\)

\(< =>2x=-5-1=-6\)

\(< =>x=-\frac{6}{2}=-3\)

Bài làm:

a) \(\left(2x+1\right)^3=-125\)

\(\Leftrightarrow\left(2x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow2x+1=-5\)

\(\Leftrightarrow2x=-6\)

\(\Rightarrow x=-3\)

b) \(\left(7-x\right)^2-\left(-11\right)=15\)

\(\Leftrightarrow\left(7-x\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=9\end{cases}}\)

c) \(2020^x-2019=-2019\)

\(\Leftrightarrow2020^x=0\)

=> ko tồn tại x thỏa mãn PT

\(\left(7-x\right)^2-\left(-11\right)=15\)

\(< =>\left(7-x\right)^2=15-11=4\)

\(< =>\left(7-x\right)^2=2^2=\left(-2\right)^2\)

\(< =>\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\x=9\end{cases}}\)

Ta có: (2x +1)³ =125

(2x +1)³ = 5³

2x + 1 = 5

2x = 4

x = 2

( 2x + 1 ) \(^3\)= 125 

( 2x + 1 )      = \(\sqrt[3]{125}\)

( 2x + 1 )      = 5

 2x               = 5 - 1

 2x               = 4

  x                = 4 : 2

  x                 = 2

(2x+1)³=125=>2x+1=5=>2x=4=>x=2

\(\hept{\begin{cases}3|x|+15=0\\16-x^2=0\\x^3+125=0\end{cases}}\)

\(\hept{\begin{cases}3x+15=0\\16-x^2=0\\x^3+125=0\end{cases}}\)

\(\hept{\begin{cases}3x=0-15\\x^2=16-0\\x^3=0-125\end{cases}}\)

\(\hept{\begin{cases}3x=-15\\x^2=16\\x^3=-125\end{cases}}\)

\(\hept{\begin{cases}x=\left(-15\right):3\\x^2=4^2=\left(-4\right)^2\\x^3=-5^3\end{cases}}\)

\(\hept{\begin{cases}x=-5\\x=4=-4\\x=-5\end{cases}}\)

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

a) 2^x . 4 = 128

2^x = 128 : 4 = 32

2^x = 2⁵

=> x = 5

x¹⁵ = x

=> x = 0 hoặc x = 1 hoặc x = -1

c) (2x + 1)³ = 125 = 5³

2x + 1 = 5

2x = 4

x = 2

d) (x-5)⁴ = (x-5)⁶

Th1: x - 5 = 0

=> x = 5

th2: x - 5 = 1

x = 6

Vậy x = 5 hoặc 6

a    2^x=128:4=32

x=5

b     x=1

c   (2x+1)=5      2x=5-1  2x=4 x=2

15×X=225-45=180=>X=12

A) X bằng 12.

B)X bằng 98.

Tk nha bạn và học tốt.