\(\left(3x-2^4\right).7^3=2\)

\(\Leftrightarrow\left(3x-16\right).343=2\)

\(\Leftrightarrow1029x-5488=2\)

\(\Leftrightarrow1029x=5490\)

\(\Leftrightarrow x=\frac{1830}{343}\)

\(\left|x-5\right|=7\)

\(\Leftrightarrow x-5=7\)

\(\Leftrightarrow x=7+5\)

\(\Leftrightarrow x=12\)

#hoc_tot#

=> x-5=7hoặc=-7

th1:x-5=7

x=7+5

x=12

th2:x-5=-7

x=-7+5

x=-2

=>x=12hoặc-2

\(\text{ ( 2x - 4 ) . ( 3 - x ) = 0 }\)

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

\(\Rightarrow x\in\left\{2;3\right\}\)

chúc bạn học tốt !!

\(4.5^5-32:2^4\)

\(=4.5^5-2^5:2^4\)

\(=4.5^5-2^1\)

\(=4.3125-2\)

\(=12500-2\)

\(=12498\)

học tốt

4.5⁵-32:2⁴= 4. 5 . 5⁴-2⁵ : 2⁴

                 = 20 . 5⁴ -2= 20.  625-2=12500-2=12498

Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)

\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)

b) 3x - 7.(5x-1) = 6 - 2.(4-3x)

=> 3x - 35x + 7 = 6 - 8 + 6x

=> 3x - 35x - 6x = 6-8 -7

-38x = -9

x = 9/38

Ta có: \(\left(x-3\right)^2=25\)

\(\Leftrightarrow\left(x-3\right)^2=5^2\left(x\inℕ^∗\right)\)

\(\Leftrightarrow x-3=5\)

\(\Leftrightarrow x=5+3\)

\(\Leftrightarrow x=8\)

hok tốt!!

(x-3)² = 25

(x-3)² = 5²

x-3 = 5

x= 5+3

x=8

\(B=\frac{2018+2019}{2019+2020}\)

\(\Rightarrow B=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)

\(\Rightarrow B< \frac{2018}{2019}+\frac{2019}{2020}=A\)

Vậy B < A

\(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(\Rightarrow B< \frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}=A\)

Vậy B < A