3x+1+3x=36
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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
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
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.
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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+1 + 3x = 36
3x . 3 + 3x . 1 = 36
3x . ( 3 + 1 ) = 36
3x . 4 = 36
3x = 36 : 4
3x = 9
3x = 32
=> x = 2