36+5x=81
tìm x
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\(\frac{x+3}{12}+\frac{2x+46}{16}=\frac{3x+9}{36}+\frac{5x+81}{6}\Leftrightarrow\frac{2\left(x+3\right)}{24}+\frac{\frac{3}{2}\left(2x+46\right)}{\frac{3}{2}.16}=\frac{\frac{2}{3}\left(3x+9\right)}{\frac{2}{3}.36}+\frac{4\left(5x+81\right)}{4.6}\)
\(\Leftrightarrow\frac{2\left(x+3\right)}{24}+\frac{3x+69}{24}=\frac{2x+6}{24}+\frac{20x+324}{24}\)\(\Leftrightarrow2x+6+3x+69=2x+6+20x+324\Leftrightarrow17x=-255\Leftrightarrow x=-\frac{255}{17}\)Vậy \(x=-\frac{255}{17}\)
\(\begin{array}{l}a){\rm{ }}3{x^2}-{\rm{ }}3x\left( {x{\rm{ }}-{\rm{ }}2} \right){\rm{ }} = {\rm{ }}36\\ \Leftrightarrow 3{x^2}-{\rm{ [}}3x.x + 3x.( - 2)] = 36\\ \Leftrightarrow 3{x^2} - (3{x^2} - 6x) = 36\\ \Leftrightarrow 3{x^2} - 3{x^2} + 6x = 36\\ \Leftrightarrow 6x = 36\\ \Leftrightarrow x = 36:6\\ \Leftrightarrow x = 6\end{array}\)
Vậy x = 6
\(\begin{array}{l}b){\rm{ }}5x\left( {4{x^2}-{\rm{ }}2x{\rm{ }} + {\rm{ }}1} \right){\rm{ }}-{\rm{ }}2x\left( {10{x^2}-{\rm{ }}5x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }} - 36\\ \Leftrightarrow 5x.4{x^2} + 5x.( - 2x) + 5x.1 - [2x.10{x^2} + 2x.( - 5x) + 2x.2] = - 36\\ \Leftrightarrow 20{x^3} - 10{x^2} + 5x - (20{x^3} - 10{x^2} + 4x) = - 36\\ \Leftrightarrow 20{x^3} - 10{x^2} + 5x - 20{x^3} + 10{x^2} - 4x = - 36\\ \Leftrightarrow (20{x^3} - 20{x^3}) + ( - 10{x^2} + 10{x^2}) + (5x - 4x) = - 36\\ \Leftrightarrow x = - 36\end{array}\)
Vậy x = -36
36+5x=81
5x=81-36
5x=45
x=45:5
x=9
5x là 5 x xờ