(x + 4x). 25 = 125
2 (x + 3 ) + 3( x + 1 ) = 34
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KV
27 tháng 7 2023
a) (125 - x) + 25 = 98
125 - x = 98 - 25
125 - x = 73
x = 125 - 73
x = 52
b) 100 : (x + 20) = 68 : 34
100 : (x + 20) = 2
x + 20 = 100 : 2
x + 20 = 50
x = 50 - 20
x = 30
c) (3x + 12) . 5 = 25 . 4 - 10
(3x + 12) . 5 = 100 - 10
(3x + 12) . 5 = 90
3x + 12 = 90 : 5
3x + 12 = 18
3x = 18 - 12
3x = 6
x = 6 : 2
x = 3
d) 210 : (2x - 3) - 20 = 10
210 : (2x - 3) = 10 + 20
210 : (2x - 3) = 30
2x - 3 = 210 : 30
2x - 3 = 70
2x = 70 + 3
2x = 73
x = 73/2
NM
27 tháng 7 2023
a, ( 125 - x ) + 25 = 98
⇒ 125 - x = 73
⇒ x = 52.
Vậy..
b, 100 : ( x + 20 ) = 68 : 34
⇒ 100 : ( x + 20 ) = 2
⇒ x + 20 = 50
⇒ x = 30
Vậy...
c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10
⇒ ( 3 . x + 12 ) . 5 = 90
⇒ 3 .x + 12 = 18
⇒ 3x = 6
⇒ x = 2.
Vậy...
d, 210 : ( 2 .x - 3 ) - 20 = 10
⇒ ( 2 .x - 3 ) - 20 = 21
⇒ 2x - 3 = 41
⇒ 2x = 44
⇒ x = 22.
Vậy..
16 tháng 2 2021
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
27 tháng 3 2020
a,-2x -(x-17)=34-(-x+25)
-2x-x+17=34+x-25
-3x+17=9+x
-3x-x=9-17
-4x=-8
-->4x=8
x=8:4
x=2
Vậy x=2
b,17-(16x-37)=2x+43
17-16x+37=2x+43
20-16x=2x+43
-16x-2x=43-20
-18x=23
x=23:(-18)
x=23/-18
Mà x là số nguyên nên --> x thuộc tập rỗng
c,-2x-3.(x-17)=34-2(-x+25)
-2x-3x+51=34-2.(-x)-25
-5x+51=9-(-2).x
-5x+(-2).x=9-51
-7x=-42
7x=42
x=42:7
x=6
Vậy x=6
13 tháng 6 2022
a: =17-43=-26
b: =-34+34+11-11+105=105
c: \(=64+125\cdot100=12564\)
d: \(=60+\left[7^3-343\right]\cdot2017^{2018}=60\)
10 tháng 4 2022
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
14 tháng 7 2023
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
NN
14 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
a) (x+4x).25=125
=> 5x = 5
=> x = 1
b) 2(x+3) + 3(x+1) = 34
=> 2x+6+3x+3 = 34
=> 5x+9=34
=> 5x=25
=> x =5
\(\left(x+4x\right).25=125\\ 5x.25=125\\ 5x=125:25\\ 5x=5\\ x=5:5\\ x=1\\ 2\left(x+3\right)+3\left(x+1\right)=34\\ 2x+6+3x+3=34\\ 5x+9=34\\ 5x=34-9\\ 5x=25\\ x=25:5\\ x=5.\)