`a)`\(A=167\times12-204:x\)

         \(A=2004-204:x\)

`x=5,1xx3,2+5,1xx6,7+51xx0,01`

`x=5,1xx3,2+5,1xx6,7+5,1xx0,1`

`x=5,1xx(3,2+6,7+0,1)`

`x=5,1xx10`

`x=51`

\(A=2004-204:51\)

\(A=2004-4\)

\(A=2000\)

`b)`\(A=2004-204:x=2002\)

\(\Rightarrow-204:x=-2\)

\(\Rightarrow x=102\)

Ta có :\(6=2.3;4=2^2\)

\(BCNN\left(6,4\right)=2^2.3=12\)

⇒x ∈ BC ( 6 , 4 ) = B ( 12 ) = { 0;12;24;36;48;60;...}

mà 16 ≤ x ≤50

⇒x∈{24;46;48}

vậyx ∈ {24;46;48}

a) Ta có : 24:x; 36:x; 150:x và x lớn nhất

⇒x là ƯCLN ( 24;36;150)

24=2³.3

36=2².3²

150=2.3.5²

UCLN (24;36;160)=2 . 3 = 6

\(71\dfrac{23}{45}+\left(43\dfrac{22}{45}+1\dfrac{1}{6}\right)\)

\(=\dfrac{3218}{45}+\left(\dfrac{1957}{45}+\dfrac{7}{6}\right)\)

\(=\dfrac{3218}{45}+\dfrac{1957}{45}+\dfrac{7}{6}\)

\(=115+\dfrac{7}{6}\) \(=\dfrac{697}{6}\)

x: (2x -1) . (5x+3/4) =0  đk x # 1/2

⇔ x = 0 hoăc 5x + 3/4 = 0⇒ x = -3/4: 5 = -3/20

vậy x = {-3/20; 0}

`(2/3x - 1) . (5x + 3/4) = 0`

`=>` \(\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\5x+\dfrac{3}{4}=0\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}\dfrac{2}{3}x=0+1\\5x=0-\dfrac{3}{4}\end{matrix}\right.\)

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

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

`=>` \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{20}\end{matrix}\right.\)

Vậy......

Help

no

Có: \(28=2^2.7\)

\(56=2^4.7\)

\(72=2^3.3^2\)

\(ƯCLN\left(28;56;72\right)=2^2=4\)

2x-2=x

2x=x+2

2x-x=x+2-x

x=2

đúng thì tick cho mình nhé