Gấp gấp . Trả lời cho mình với
2x . 3^x = 3^x + 2x +1
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\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
\(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
=> \(x=-\frac{13}{4}\)
=> 3.(x+2) = 2x-4
=> 3x+6 = 2x-4
=> 3x = 2x-4-6 = 2x-10
=> 10 = 2x-3x = -x
=> x = 10 : (-1) = -10
Tk mk nha
=>\(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\)
+, x+\(\frac{1}{2}\)=0 +,\(\frac{2}{3}-2x=0\)
x=\(-\frac{1}{2}\) =>\(\frac{2}{3}=2x\)
=>\(x=\frac{1}{3}\)
Vậy........
\(a,-2.\left(x+7\right)+3.\left(x-2\right)=-2\)
\(-2x-14+3x-6=-2\)
\(x-20=-2\)
\(x=-2+20\)
\(x=18\)
\(b,-7-2x=-37-\left(-26\right)\)
\(-7-2x=-11\)
\(-2x=-11+7\)
\(-2x=-4\)
\(x=2\)
\(c,\left(3x+9\right).\left(11-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\11-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=11\end{cases}}}\)
\(\orbr{\begin{cases}3x+9=0\\11-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-9\\x=11\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=11\end{cases}}\)
\(2x.3^x=3^x+2x+1\)
\(\Rightarrow2x.3^x-3^x-2x-1=0\)
\(\Rightarrow3^x.\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Rightarrow\left(3^x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow3^x-1=0\) hoặc \(2x+1=0\)
\(\Rightarrow3^x=1\) hoặc \(2x=-1\)
\(\Rightarrow x\in\left\{-\frac{1}{2};0\right\}\)