Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)

\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)

\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)

\(=\frac{1}{2}-\frac{1}{37}\)

\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)

\(=\frac{1}{2}.\frac{1}{7}+\frac{1}{7}.\frac{1}{12}...+\frac{1}{32}.\frac{1}{37}\)

= \(\frac{1}{2}+\frac{1}{37}\)

= \(\frac{37}{74}+\frac{2}{74}\)

\(\frac{39}{74}\)

tíck mik nha, mik nhanh nhất

\(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)

\(=\frac{1}{2}-\frac{1}{27}\)

\(=\frac{25}{54}\)

Ta có |5x-5012| = |-5x+5012| >_ -5x +5012.Dấu "=" khi -5x+5012>0

         | 5x+300|>_ 5x+300.Dấu "=" khi 5x+300>0

=> |-5x+5012| + |5x+300| >_ -5x+5012 +5x + 300

=> A >_ 5312

Dấu "=" khi -5x+5012>0 => x<5012/5

                   5x+300> 0   => x>-60

Vậy Min A = 5312 khi -60<x<5012/5

+) 5x - 16 = 40 + x

<=> 5x - x = 16 + 40

<=> 4x = 56

<=> x = 56 : 4

<=> x= 14

+) 3x - 6 = 5x + 2

<=> 3x - 5x = 6 + 2

<=> -2x = 8

<=> x =- 4

+) x + 15 = 20 - 4x

<=> x + 4x = 20 - 15

<=> 5x = 5

<=> x= 1 

+) -12 + x = 5x -20

<=> x - 5x = 12 - 20

<=> -4x = -8

<=> x= 2

T^T , T sai chỗ nào vậy

bạn đoán xem

bằng 2270/5049 đó bạn

\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{17}+\frac{5}{22.27}\)

\(=\frac{2270}{5049}\)

\(5x\left(x-2018\right)-x+2018=0\)

\(5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\left(x-2018\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

Vậy.........

x = 2018

x = 1/5

t i c k nha

\(5x\left(x-2018\right)-x+2018=0\)

\(\Leftrightarrow5x\left(x-2018\right)-\left(x-2018\right)=0\)

\(\Leftrightarrow\left(x-2018\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2018=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{5}\end{cases}}\)

Ta có: \(5^x\cdot5^{x-1}=125\)

\(\Leftrightarrow5^x\cdot\dfrac{5^x}{5}=125\)

\(\Leftrightarrow25^x=625\)

hay x=2