Tìm x, biết :
3x + 3x+1 + 3x+2 = 39
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Tìm x biết:
5. ( x-1 ) - 7.( x-2 ) = 2x -39
Tìm x thuộc Z biết:
x - 3 - 14.( x-2 )= -3x -3
\(3x+7⋮x-2\)
5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39
<=> 5x - 5 - 7x + 14 = 2x - 39
<=> 5x - 7x - 2x = -39 + 5 - 14
<=> -4x = -48
<=> x = 12
x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
a, 2 x ( 15 - 3x ) = 12
=> 15 - 3x = 6
=> 3x = 9
=> x = 3
b, 39 - 2 x ( 31 - 3x ) = 15
=> 2 x ( 31 - 3x ) = 24
=> 31 - 3x = 12
=> 3x = 19
=> x = \(\frac{19}{3}\)
nếu mk sửa đề có sai thì nhờ bạn sửa đề lại giúp mk nha
a) \(55-4x-4\left(-x+3\right)=6-2\left(-8-3x\right)\)
\(55-4x+4x-12=6+16+6x\)
\(43-6-16=6x\)
\(6x=21\)
\(x=3,5\)
b) \(-5\left(-2x-6\right)-9\left(4-7x\right)=51-3x+6\left(x-9\right)\)
\(10x+30-36+63x=51-3x+6x-54\)
\(73x-6=-3+3x\)
\(73x-3x=-3+6\)
\(70x=3\)
\(x=\frac{3}{70}\)
c) \(93+\left|6-3x\right|-39=231\)
\(\left|6-3x\right|+54=231\)
\(\left|6-3x\right|=177\)
\(\Rightarrow\orbr{\begin{cases}6-3x=177\\6-3x=-177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6-177\\3x=6+177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-171\\3x=183\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-57\\x=61\end{cases}}\)
\(3^{x-1}+3^x+3^{x+1}=39\)
\(3^{x-1}+3^{x-1}.3+9.3^{x-1}=39\)
\(13.3^{x-1}=39\)
\(3^{x-1}=39:13=3\)
\(x-1=1\)
\(x=2\)
Sửa đề: 3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹ . 13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
\(\Leftrightarrow3^{x-1}\left(1+3+3^2\right)=39\\ \Leftrightarrow3^{x-1}\cdot13=39\\ \Leftrightarrow3^{x-1}=3=3^1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\)
\(\Leftrightarrow3^x\cdot\dfrac{13}{3}=39\)
\(\Leftrightarrow x=2\)
a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow9x^2-3x-9x^2+1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b: Ta có: \(x^2-5x+25-5x=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
b: \(3x^2-2x-1=0\)
=>\(3x^2-3x+x-1=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: Bạn ghi lại đề đi bạn
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`