= 7 x ( 2 x230) + 7 x 540

= 7 x 460 + 7 x 540

= 7 x ( 460 + 540)

= 7 x 1000

= 7000

\(x+x+x+x-x=540\)

\(x+x+x+0=540\)

\(x\cdot3=540\)

\(x=540:3\)

\(x=180\)

X+X+X=540

Xx4    =540

X        = 540:4

X        =135

mh làm sai rồi

X=180 mới đúng

x+x+x+x-x=540

x+x+x=540

\(3\times x=540\)

x=540:3

x=180

180 bạn nhé!

xcvdzssssssssssssssssssssssssssssssssssssssssssssssssser

\(34\cdot x-3\cdot x-x=540\)

\(\left(34-3-1\right)\cdot x=540\)

\(30\cdot x=540\)

\(x=540:30\)

\(x=18\)

34 x X - 3 x X - X = 540

( 34 - 3 - 1 ) x X = 540

30 x X = 540

X = 540 : 30

X = 18

34 × x - 3 × x - x = 540 

34 ×  - 3 × x - x × 1 = 540 

( 34 - 3 - 1 ) × x = 540 

30 × x = 540 

x = 540 : 30 

x = 18 

Vậy x = 18

2 * x + 3 * x +  4 *  x + 5 * x + 6 * x + 7 * x = 540

x * ( 2 + 3 + 4 + 5 + 6 + 7 + 8 ) = 540

x * 27 = 540

x = 540 : 27

x = 20 

Vậy x = 20

Chúc bạn học tốt!

2 * x + 3 * x +  4 *  x + 5 * x + 6 * x + 7 * x = 540

x * ( 2 + 3 + 4 + 5 + 6 + 7 + 8 ) = 540

x * 27 = 540

x = 540 : 27

x = 20 

Vậy x = 20

Chúc bạn học tốt!

2*x+3*x4*x+5*x+6*x+7*x=540

x*(2+3+4+5+6+7)=540

x*27=540

   x=540/27

   x=20

Cách b thuận tiện nhất

Lời giải:

\(\frac{20,2\times 5,1-30,3\times 3,4+14,58}{14,58\times 460+7,29\times 540\times 2}=\frac{10,1\times 2\times 5,1-10,1\times 3\times 3,4+14,58}{14,58\times 460+14,58\times 540}\\ =\frac{10,1\times 10,2-10,1\times 10,2+14,58}{14,58\times (460+540)}\\ =\frac{14,58}{14,58\times 1000}=\frac{1}{1000}\)

\(\left[{}\begin{matrix}x.36=540\Rightarrow x=540:36\Rightarrow x=15\\2040:x=85\Rightarrow x=2040:85\Rightarrow x=24\\x:53=60\Rightarrow x=60.53\Rightarrow x=3180\end{matrix}\right.\)

a: \(\Leftrightarrow x=15\)

b: x=24

X x 36= 540        X : 53 = 60
X=540 : 36          X= 60 x 53
X= 15.                 X= 3180

2040 : X = 85

X = 2040 : 85

X=24
 

\(7\cdot3^{x-1}-3^{x+2}=540\)

=>\(7\cdot3^x\cdot\dfrac{1}{3}-3^x\cdot9=540\)

=>\(3^x\left(\dfrac{7}{3}-9\right)=540\)

=>\(3^x\cdot\dfrac{-20}{3}=540\)

=>\(3^x=-540:\dfrac{20}{3}=-540\cdot\dfrac{3}{20}=-27\cdot3=-81\)

=>\(x=log_3\left(-81\right)\)