Tìm x :
a) 10-2x=25-3x
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Ta có: 10 - 2x = 25 - 3x
<=> 10 - 2x - 25 + 3x = 0
<=> x - 15 = 0
=> x = 15
10 - 2x = 25 - 3x
10 - 25 = 2x - 3x
-15 = -x
=> x = 15
10-2x=25-3x
=>2x+3x=25-10
=>x(2+3)=15
=>5x=15
=>x=15:5
=>x=3
a, 70 - 5 ( x - 3 ) = 45
⇔ 5 . x - 3 = 70 - 45
⇔ 5 x - 3 = 35
⇔ x - 3 = 35 : 5 ⇔ x - 3 = 7
⇔ x = 10
b, 10 + 2 x = 4 5 : 4 3
⇔ 10 + 2 x = 4 2 ⇔ 10 + 2 x = 16
⇔ 2 x = 4 ⇔ x = 2
c, 60 - 3 x - 2 = 51
⇔ 3 x - 2 = 60 - 51
⇔ 3 x - 2 = 9
⇔ x - 2 = 3 ⇔ x = 5
d, 4 x - 20 = 2 5 : 2 3
⇔ 4 x - 20 = 2 2 ⇔ 4 x - 20 = 4
⇔ 4 x = 24 ⇔ x = 6
Tách ?
`a, 70 -5.(x-3) =45`
`=> 5.(x-3)= 70-45`
`=> 5.(x-3)=25`
`=>x-3=25:5`
`=>x-3=5`
`=>x= 5+3`
`=>x=8`
______
`b,10 + 2.x = 4^5:4^3`
`=> 10 + 2.x = 4^(5-3)`
`=> 10 + 2.x =4^2=16`
`=> 2.x=16-10`
`=>2.x=6`
`=>x=6:2`
`=>x=3`
_____
`c,60-3.x-2=51`
`=> 60-3.x= 51+2`
`=> 60-3.x=53`
`=>3.x=60-53`
`=> 3.x= 7`
`=>x= 7/3`
____
`d, 4.x-20=2^5:2^3`
`=> 4.x-20=2^(5-3)`
`=> 4.x-20=2^2`
`=> 4.x= 4+20`
`=>4.x=24`
`=>x=24:4`
`=>x=6`
____
`2^x . 4=16`
`=> 2^x=16:4`
`=>2^x= 4`
`=>2^x=2^2`
`=>x=2`
____
`f, 3^x . 3=243`
`=>3^x=243:3`
`=> 3^x=81`
`=> 3^x=3^3`
`=>x=3`
_____
`g, 64. 4^x =16^8`
`=> 4^3 . 4^x=(4^2)^8`
`=> 4^3 . 4^x = 4^(16)`
`=> 4^x= 4^(16-3)`
`=>4^x=4^(13)`
`=>x=13`
_____
`2^x . 16^2 =1024`
`=> 2^x= 1024 : 16^2`
`=>2^x=4`
`=>2^x=2^2`
`=>x=2`
a: =>5(x-3)=25
=>x-3=5
=>x=8
b: =>2x=16-10=6
=>x=3
c: =>58-3x=51
=>3x=7
=>x=7/3
d: =>4x-20=4
=>4x=24
=>x=6
e: =>2^x=4
=>2^x=2^2
=>x=2
f: =>3^x=81
=>3^x=3^4
=>x=4
g: =>4^x*4^3=4^16
=>x+3=16
=>x=13
h: =>2^x=1024/256=4=2^2
=>x=2
25-{14-[(2-x)-(x+13)+23]}=62-{8-[(2x-10)-(29-3x)+8]}
<=>25-14+2-x-x-13+23=62-8+2x-10-29+3x+8
<=>13-x-x-13+23=2x+3x+8-10-29
<=>-x-x+23=2x+3x+(-31)
<=>23-(x+x)=2x+2x-31
<=>23-2x=2x+3x-31
<=>23+31=2x+2x+3x
<=>54=x(2+2+3)
<=>54=x.7
=>x=54/7
Vậy x=54/7
<=>54=x.7
`#040911`
`a)`
`(2x - 1)^2 - (2x + 5)(2x + 1) = 10`
`\Leftrightarrow 4x^2 - 4x + 1 - (4x^2 + 12x + 5) = 10`
`\Leftrightarrow 4x^2 - 4x + 1 - 4x^2 - 12x - 5 = 10`
`\Leftrightarrow (4x^2 - 4x^2) - (4x + 12x) + (1 - 5) = 10`
`\Leftrightarrow -16x - 4 = 10`
`\Leftrightarrow -16x = 10 + 4`
`\Leftrightarrow -16x = 14`
`\Leftrightarrow x = \dfrac{-7}{8}`
Vậy, `x= \dfrac{-7}{8}`
`b)`
`9^2(x - 1) + 25(1 - x) = 0`
`\Leftrightarrow 9^2(x - 1) - 25(x - 1) = 0`
`\Leftrightarrow (x - 1)(9^2 - 25) = 0`
`\Leftrightarrow`\(\left[{}\begin{matrix}x-1=0\\9^2-5^2=0\end{matrix}\right.\)
`\Leftrightarrow`\(\left[{}\begin{matrix}x=1\\\left(9-5\right)\left(9+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\4\cdot14=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\56=0\left(\text{vô lý}\right)\end{matrix}\right.\\ \text{Vậy, x = 1}\)
`c)`
\(x^2+3x-4=0\)
`\Leftrightarrow x^2 + 4x - x - 4 = 0`
`\Leftrightarrow (x^2 - x) + (4x - 4) = 0`
`\Leftrightarrow x(x - 1) + 4(x - 1) = 0`
`\Leftrightarrow (x + 4)(x - 1) = 0`
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\\ \text{ Vậy, }x\in\left(-4;1\right)\)
10+25=3x+2x
35=5x
x=7
a) 10-2x=25-3x
=10-25=2x-3x
= -15 = -x
Vậy x=15