Tìm x, biết :
3x + 3x+1 + 3x+2 = 39
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\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
HTDT-sai-rùi
x-3-14(x-2)=-3x-3
x-3+28-14x=-3x-3
-13x+35=-3x-3
-13x+3x=-3-35
-10x=-38
x=38/10=19/5
Tìm stn x,biết: a, 2 x ( 15 - 3x ) = 12 b, 39 - 2 x ( 31 - 3x ) - 15
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
\(2\left(15-3x\right)\)\(=12\)
\(15-3x=12:2=6\)
\(3x=15-6=9\)
\(x=9:3=3\)
\(b,39-2\left(31-3x\right)\)\(-15\)
Đến đây cho anh hỏi thế biểu thức này giá trị bằng bao nhiêu
bài 1:tìm x , biết
a,55-4x-4(-x+3)=6-2(-8-3x)
b,-5(-2x-6)-9(4-7x)=51-3x+6(x-9)
c,93+/6-3x/-39=231
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) \(93+\left|6-3x\right|-39=231\)
\(\left|6-3x\right|+54=231\)
\(\left|6-3x\right|=177\)
\(\Rightarrow\orbr{\begin{cases}6-3x=177\\6-3x=-177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6-177\\3x=6+177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-171\\3x=183\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-57\\x=61\end{cases}}\)
a) \(55-4x-4\left(-x+3\right)=6-2\left(-8-3x\right)\)
\(\Leftrightarrow\)\(55-4x+4x-12=6+16+6x\)
\(\Leftrightarrow\)\(6x+22=43\)
\(\Leftrightarrow\)\(6x=43-22=21\)
\(\Leftrightarrow\)\(x=21:6=3.5\)
Vậy...
b)pt\(\Leftrightarrow10x+30-36+63x=51-3x+6x-54\)
tự tiếp
tìm x:
3x-1 + 3x + 3x+1 = 39
giúp mik với ;-;
\(3^{x-1}+3^x+3^{x+1}=39\)
\(3^{x-1}+3^{x-1}.3+9.3^{x-1}=39\)
\(13.3^{x-1}=39\)
\(3^{x-1}=39:13=3\)
\(x-1=1\)
\(x=2\)
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
Tìm x; biết: 3x + 3x+1 + 3x+2 + 3x+3 = 1080
đặt 3x ra ngoài bạn nhé bàn phím mik hỏng rồi ;-;
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=1080\)
\(\Rightarrow3^x+3^x.3^1+3^x.3^2+3^x.3^3=1080\)
\(3^x.\left(1+3+9+27\right)=1080\)
\(3^x.40=1080\)
\(3^x=27\)
\(3^x=3^3\)
Vậy: x = 3
Giúp mình con tính này với:
3x-1 + 3x + 3x+1 = 39 Bài tìm x nha
\(\Leftrightarrow3^{x-1}\left(1+3+3^2\right)=39\\ \Leftrightarrow3^{x-1}\cdot13=39\\ \Leftrightarrow3^{x-1}=3=3^1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\)
\(\Leftrightarrow3^x\cdot\dfrac{13}{3}=39\)
\(\Leftrightarrow x=2\)
\(3^{x-1}+3^x+3^{x+1}=39\\ \Rightarrow3^x:3+3^x+3^x.3=39\\ \Rightarrow3^x.\dfrac{1}{3}+3^x+3^x.3=39\\ \Rightarrow3^x\left(\dfrac{1}{3}+1+3\right)=39\\ \Rightarrow3^x.\dfrac{13}{3}=39\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2\)
TÌM X BIẾT :
a/ 3x ( 3x -1 ) - ( 3x + 1 ) ( 3x - 1 ) = 0
b/ \(x^2\) - 5x + 25 - 5x = 0
KHÔNG BỎ BƯỚC Ạ !
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
tìm x biết (3x + 4)^2 - ( 3x -1).(3x+1)
\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=9x^2+24x+16-9x^2+1=24x+17\)
Đặt \(24x+17=0\Leftrightarrow x=-\frac{17}{24}\)
Tìm x biết a) 3x^2+x)4-3x)=12 b)3x^2-2x-1=0
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
Bài 2: Tìm x, biết:
a/ 12x(x – 5) – 3x(4x - 10) = 120
b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)
c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)
$ 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`