393-(126+3X)=0

        126+3X.  =393-0

          126+3X =393

                   3X =393-126

                   3X =267

                   X.   =267 : 3

                  X.     =89

Còn câu dưới hình như đề sai

(3x+1)^3=-216=(-6)^3

=>3x+1=-6

=>x=...

3^x-1+5.3^x-1=162

=>3^x-1.(1+5)=162

=>3^x-1.6=162

=>3^x-1=162:6=27

=>3^x-1=3^3

=>x-1=3

=>x=4

c)
\(x^3-3.x^2.6+3.x.6^2-6^3=0\)
\(\left(x-6\right)^3=0\)
x-6=0
x=6
d)
\(x^3-3.x^2.1+3.x.1^2-1-x^3-3x-2=0\)
\(x^3-3x^2+3x-1-x^3-3x^2-2=0\)
\(-6x^2-3=0\)
\(-3\left(2x^2+1\right)=0\)
\(2x^2+1=0\)
2x²=-1
x²=1/2
x=\(\dfrac{\sqrt{2}}{2}\)

\(x=0,\)biểu thức không thỏa mãn.

\(x\ge1\) thì \(\hept{\begin{cases}3^x⋮3\\3x⋮3\end{cases}}\Rightarrow3^x-3x⋮3\)

\(\Rightarrow3^x+2-3x\)không chia hết cho 3

Mà 216 chia hết cho 3 nên vô lý.

Mình không hiểu rõ đề, bạn dùng f(x) nhé. Mình nhìn đề thành thế này.

\(3^x\times\left(3^2-1\right)=216\)

\(3^x\times\left(9-1\right)=216\)

\(3^x\times8=216\)

\(3^x=\frac{216}{8}\)

\(3^x=27\)

\(3^x=3^3\)

\(x=3\)

\(6^x=8\times3^x\)

\(\frac{6^x}{3^x}=8\)

\(\left(\frac{6}{3}\right)^x=2^3\)

\(2^x=2^3\)

\(x=3\)

\(\left(3x+1\right)^3=1\)

\(3x+1=1\)

\(3x=1-1\)

\(3x=0\)

\(x=0\)

a)\(\left(1-x\right)^3=216\)

\(\Rightarrow1-x=6\)

\(\Rightarrow x=-5\)

b)\(3^{x+1}-3^x=162\)

\(\Rightarrow3^x\left(3-1\right)=162\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

c)\(5^{x+1}-2.5^x=375\)

\(\Rightarrow5^x\left(5-2\right)=375\)

\(\Rightarrow5^x.3=375\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

a) x=-5

giải cho mk luôn đi

a: \(=\dfrac{2\left(\sqrt{3}-2\right)}{2\left(\sqrt{2}-1\right)}-6\sqrt{2}\)

\(=\dfrac{-2+\sqrt{3}}{\sqrt{2}-1}-6\sqrt{2}\)

\(=\left(-2+\sqrt{3}\right)\left(\sqrt{2}+1\right)-6\sqrt{2}\)

\(=-2\sqrt{2}-2+\sqrt{6}+\sqrt{3}-6\sqrt{2}\)

\(=-8\sqrt{2}+\sqrt{6}+\sqrt{3}-2\)

b: \(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

c:\(=\dfrac{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)}{1+\sqrt{y}}=1+\sqrt{x}\)

d: \(=\dfrac{\left(x-\sqrt{3x}+3\right)}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}+\sqrt{3}}\)