Ta có:

\(3^{x+1}+3^x=36\\ \Leftrightarrow3^x.4=36\\ \Leftrightarrow3^x=9\\ \Leftrightarrow x=2\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

X bằng 1

=>3x+1=4

hay x=1

x=1

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)

b) \(\left(3x+1\right)^3=-27\)

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

\(\Rightarrow3x+1=-3\)

\(\Rightarrow3x=\left(-3\right)-1\)

\(\Rightarrow3x=-4\)

\(\Rightarrow x=\left(-4\right):3\)

\(\Rightarrow x=-\frac{4}{3}\)

Vậy \(x=-\frac{4}{3}.\)

Mấy câu sau làm tương tự nhé.

Chúc bạn học tốt!

c)\(\left(3x-2\right)^2=36\\ \Leftrightarrow\left(3x-2\right)^2=\left(\pm6\right)^2\\ \Rightarrow\left\{{}\begin{matrix}3x-2=6\\3x-2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{4}{3}\end{matrix}\right.\)

d)\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\\ \Leftrightarrow\left(\frac{2}{5}-3x\right)^2=\left(\pm\frac{3}{5}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\frac{2}{5}-3x=\frac{3}{5}\\\frac{2}{5}-3x=-\frac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{15}\\x=\frac{1}{3}\end{matrix}\right.\)

\(\left(2-3x\right)^2=\dfrac{1}{36}=\left(\dfrac{1}{6}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=\dfrac{1}{6}\\2-3x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{18}\\x=\dfrac{13}{18}\end{matrix}\right.\)

\(\left[{}\begin{matrix}2-3x=\dfrac{1}{6}\\2-3x=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{11}{6}\\3x=\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{6}:3=\dfrac{11}{18}\\x=\dfrac{13}{6}:3=\dfrac{13}{18}\end{matrix}\right.\)

\(3x^2-3x\left(-2+x\right)=36\)

\(\Leftrightarrow3x^2+6x-3x^2-36=0\)

\(\Leftrightarrow6x-36=0\)

\(\Leftrightarrow x=6\)