\(2.3^x=162\)

\(3^x=162:2\)

\(3^x=81\)

\(\Rightarrow x=4\)

Vậy x = 4

theo đề bài ta có : 

2.3^x = 162 

3^x = 162 :2 

3^x    = 81

3^x    = 3⁴

x = 4

\(2.3^x=162\)

\(3^x=162:2\)

\(3^x=81\)

\(3^x=3^4\)

Vậy x=4

Ta có: \(\dfrac{x}{3}=\dfrac{y}{6}\Rightarrow x=\dfrac{3y}{6}=\dfrac{1}{2}y\)

Theo đề bài ta có : \(xy=162\Rightarrow\dfrac{1}{2}y.y=162\Rightarrow y^2=324\Rightarrow y=18\)

\(\Rightarrow x=\dfrac{1}{2}y=9\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=6k\end{matrix}\right.\)

Ta có: xy=162

\(\Leftrightarrow18k^2=162\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=9\\y=6k=18\end{matrix}\right.\)

Trường hợp 2: k=-3

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-9\\y=6k=-18\end{matrix}\right.\)

327 x (162 + x) = 67035

            162 + x  = 67035 : 327

            162 + x  = 205

                      x  = 205 - 162

                      x  = 43

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\(327\times\left(162+x\right)=67035\)

\(162+x=67035\div327\)

\(162+x=205\)

\(x=205-162\)

\(x=43\)

\(327\times\left(162+x\right)=67035\)

                 \(162+x=67035:327\)

                 \(162+x=\text{205}\)

                               \(x=205-162\)

                               \(x=\text{43}\)

a) \(\left(x-7\right)-\left(5-x\right)=12-\left(-8+x\right)\)

\(\Leftrightarrow x-7-5+x=12+8-x\)

\(\Leftrightarrow2x-12=20-x\)

\(\Leftrightarrow2x+x=12+20\)

\(\Leftrightarrow3x=32\)

\(\Leftrightarrow x=\frac{32}{3}\)

b) 124 + ( 13 - 16 ) = 162 - ( x + 162 )

<=> 124 - 3 = 162 - x - 162

<=> 121 = -x

<=> x = -121