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26 tháng 12 2021

\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)

Từ đó bạn tìm những giá trị của x nha!

10 tháng 12 2021

b: \(\Leftrightarrow8x=41\)

hay x=41/8

7 tháng 2 2019

a) -21 + (4 - x) = -17 + (-20) + 5

=>-21 + 4 - x = -32

=> -17 - x = -32

=> x = -17 + 32

=> x = 15

b) (15 - x) - (+9) = 34 - (-31)

=> 15 - x - 9 = 34 + 31

=> 6 - x = 65

=> x = 6 - 65

=> x = -59

7 tháng 2 2019

c) (17 + x) - (-12) = -14 - (-10)

=> 17 + x + 12 = -14 + 10

=> 29 + x = -4

=>  x = -4 - 29

=> x = -33

c) (x + 24) + 15 = 8 - 17

=> x + 24  + 15 = -9

=> x + 39 = -9

=> x = -9 - 39

=> x = -48

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

2 tháng 8 2023

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

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

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

 

 

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

 

4 tháng 10 2021

a) \(\left(x-17\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-15\end{matrix}\right.\)

b) \(\left(6-x\right)\left(x-35\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=35\end{matrix}\right.\)

a) \(\left|x\right|+\left|-23\right|=\left|17\right|\)

\(\left|x\right|+23=17\)

\(\left|x\right|=17-23\)

\(\left|x\right|=-6\)

=> x ko có giá trị

b) \(\left|-5\right|.\left|x\right|+20\)

sai đề

10 tháng 1 2022

\(\text{(−1)+(−3)+...+(−199)+(−201)(−1)+(−3)+...+(−199)+(−201)}\)

=\(\text{−(1+3+...+199+201)=−(1+3+...+199+201)}\)

=\(\dfrac{\left(201+1\right).\left[\left(201-1\right)\right]:2+1}{2}\)

\(\dfrac{-200.102}{2}=\dfrac{-20400}{2}=-10200\)

\(\text{17 + ( − 20 ) + 23 + ( − 26 ) + . . . + 53 + ( − 56 ) = [ 17 + ( − 20 ) ] + [ 23 + ( − 26 ) ] + . . . + [ 53 + ( − 56 ) ] = ( − 3 ) + ( − 3 ) + . . . + ( − 3 ) = ( − 3 ) . ( 7 ) = − 21}\)

\(\text{=17 + ( − 20 ) + 23 + ( − 26 ) + . . . + 53 + ( − 56 ) = [ 17 + ( − 20 ) ] + [ 23 + ( − 26 ) ] + . . . + [ 53 + ( − 56 ) ] = ( − 3 ) + ( − 3 ) + . . . + ( − 3 ) = ( − 3 ) . ( 7 ) = − 21}\)

 \(\text{ = ( − 3 ) + ( − 3 ) + . . . + ( − 3 )}\)

\(\text{= ( − 3 ) . ( 7 ) = − 21}\)