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R(x) = 2x2 + 3x - 1
- M(x) = -x3 + x2
x3 + x2 + 3x - 1
Vậy R(x) - M(x) = x3 + x2 + 3x - 1
a. Kiểm tra lại mẫu số vế phải, \(7-5x\) hay \(7-3x\)
b. ĐKXĐ: \(x\ne-\dfrac{5}{3}\)
\(\dfrac{3x+5}{12}=\dfrac{3}{5+3x}\)
\(\Leftrightarrow\dfrac{\left(3x+5\right)^2}{12\left(3x+5\right)}=\dfrac{36}{12\left(3x+5\right)}\)
\(\Rightarrow\left(3x+5\right)^2=36=6^2\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=6\\3x+5=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\) (thỏa mãn)
a: Sửa đề: 6x^3
\(\dfrac{6x^3-4x^2+3x-2}{3x-2}=\dfrac{2x^2\left(3x-2\right)+3x-2}{3x-2}=2x^2+1\)
b: \(\dfrac{6x^3-3x^2+4x+2}{3x^2+2}\)
\(=\dfrac{6x^3+4x-3x^2-2+4}{3x^2+2}\)
\(=2x-1+\dfrac{4}{3x^2+2}\)
\(\Leftrightarrow\frac{3x+1}{17}+\frac{3x+1}{19}+\frac{3x+1}{23}-\frac{3x+1}{29}-\frac{3x+1}{31}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{23}-\frac{1}{29}-\frac{1}{31}\right)=0\)
\(\Leftrightarrow3x+1=0\) ( vì \(\frac{1}{17}+\frac{1}{19}+\frac{1}{23}-\frac{1}{29}-\frac{1}{31}\ne0\))
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
\(\frac{3x+1}{17}+\frac{3x+1}{19}+\frac{3x+1}{23}=\frac{3x+1}{29}+\frac{3x+1}{31}\)
\(\Rightarrow\frac{3x+1}{17}+\frac{3x+1}{19}+\frac{3x+1}{23}-\frac{3x+1}{29}-\frac{3x+1}{31}=0\)
\(\Rightarrow\left(3x+1\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{23}-\frac{1}{29}-\frac{1}{31}\right)=0\)
Mà \(\frac{1}{17}+\frac{1}{19}+\frac{1}{23}-\frac{1}{29}-\frac{1}{31}\ne0\)
\(\Rightarrow3x+1=0\)
\(\Rightarrow x=-\frac{1}{3}\)
=3x(16x+1)
\(48x^2+3x=3x\left(16x+1\right)\)