Bài 2:
a) (x+1)(2x-3)-3(x-2)
=2(x-1)\(^2\)
b) (x+1)(x\(^2\)-x+1)-2x
=x(x-1)(x+1)
c) \(\dfrac{x}{3}\)-\(\dfrac{5x}{6}\)-\(\dfrac{15x}{12}\)=\(\dfrac{x}{4}\)-5
d) \(\dfrac{x-1}{2}\)-\(\dfrac{x+1}{15}\)-
\(\dfrac{2x-13}{6}\)=0
e) \(\dfrac{3\left(5x-2\right)}{4}\)-2
=\(\dfrac{7x}{3}\)-5(x-7)
g) \(\dfrac{x-3}{11}\)+\(\dfrac{x+1}{3}\)
=\(\dfrac{x+7}{9}\)-1
h) \(\dfrac{3x-0,4}{2}\)+\(\dfrac{1,5-2x}{3}\)
=\(\dfrac{x+0,5}{5}\)
a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)
\(\Leftrightarrow1=0\)(vô lý)
Vậy: \(S=\varnothing\)
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