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30 tháng 8 2023

Em ghi đề đầy đủ, chính xác lại!

30 tháng 8 2023

Bạn xem lại đề bài

5 tháng 6 2017

a, Đặt A = \(\frac{9}{5.32}+\frac{9}{8.44}+\frac{9}{11.56}+\frac{9}{14.68}+\frac{9}{17.80}\)

\(=\frac{1}{4}\left(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+\frac{9}{17.20}\right)\)

\(=\frac{3}{4}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{3}{4}\cdot\frac{3}{20}=\frac{9}{80}\)

b, Đặt B = \(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}\)

\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{6}+\frac{1}{7}\right)\)

\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}=\frac{6}{7}\)

29 tháng 11 2023

\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)

=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)

=>\(156\cdot5^x=44\cdot89-16=3900\)

=>\(5^x=\dfrac{3900}{156}=25\)

=>x=2

AH
Akai Haruma
Giáo viên
2 tháng 12 2023

Lời giải:

$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$

$5^x(1+5+5^2+5^3)=88.89:2-16$

$5^x.156=3900$

$5^x=3900:156=25=5^2$

$\Rightarrow x=2$

30 tháng 8 2023

2⁵ˣ⁺¹ - 2⁵ˣ = 32

2⁵ˣ.(2 - 1) = 2⁵

2⁵ˣ = 2⁵

5x = 5

x = 5 : 5

x = 1

30 tháng 8 2023

\(2^{5x+1}-2^{5x}=32\)

\(\Rightarrow2^{5x+1}-2^{5x}=2^5\)

\(\Rightarrow2^{5x}\cdot2-2^{5x}\cdot1=2^5\)

\(\Rightarrow2^{5x}\cdot\left(2-1\right)=2^5\)

\(\Rightarrow2^{5x}\cdot1=2^5\)

\(\Rightarrow2^{5x}=2^5\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=\dfrac{5}{5}\)

\(\Rightarrow x=1\)

`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)

\(\Leftrightarrow3x-5-5x+10-4x-4=0\)

\(\Leftrightarrow-6x+1=0\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)

2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)

\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)

\(\Leftrightarrow-3x=-9\)

hay x=3

3) Ta có: \(3x^2+2x=0\)

\(\Leftrightarrow x\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)

4) Ta có: \(x^2-5x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) Ta có: \(\left(2x-3\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

6) Ta có: \(\left(5x-1\right)^3=125\)

\(\Leftrightarrow5x-1=5\)

\(\Leftrightarrow5x=6\)

hay \(x=\dfrac{6}{5}\)

7) Ta có: \(3^{x+1}=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

24 tháng 1

5x+5x+1+5x+2=31

5x + 5x + 5x = 31 - 2 - 1 

15x = 28

x= 28/15

6 tháng 9 2023

3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)

15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)

(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0

       27\(x^3\) + 3\(x^2\)  - 15\(x^4\) - 27\(x^3\) = 0

                     3\(x^2\) - 15\(x^4\)      = 0

                     3\(x^2\).(1 - 5\(x^2\)) = 0

                         \(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)  

                          

 

 

            

                    

 

 

                         

              

24 tháng 9 2016

a, 42x - 6 = 1

=> 42 x = 7 

=> x = 6

b, 5x + 5x + 1 +5x + 2 = 775

=> 15 x + 3 = 775

=> 15 x = 772

=> x = 772/ 15

24 tháng 9 2016

a,= > = 6

b,= > x = 772 / 15