a) Ta có: 3(2x-1)-5(x-3)+6(3x-4) = 24
\(\Leftrightarrow\) 6x-3-5x+15+18x-24 = 24
\(\Leftrightarrow\) 19x-12 = 24
\(\Leftrightarrow\) 19x = 36
\(\Leftrightarrow\) x = \(\dfrac{36}{19}\)
b) Ta có: 6x2-(2x-3)(3x+2)-1 = 0
\(\Leftrightarrow\) 6x2-6x2-4x+9x+6-1 = 0
\(\Leftrightarrow\) 5x = -5
\(\Leftrightarrow\) x = -1
c) Ta có: 4x2-12x = 9
\(\Leftrightarrow\) (2x)2-2.2x.3 + 32 = 18
\(\Leftrightarrow\) (2x-3)2 = 18
\(\Leftrightarrow\) 2x-3 = \(\sqrt{18}\)
\(\Leftrightarrow\) x = \(\dfrac{\sqrt{18}+3}{2}\)
\(a,3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=24\)
\(\Rightarrow6x-3-5x+15+18x-24=24\)
\(\Rightarrow\left(6x-5x+18x\right)+\left(-3+15-24\right)=24\)
\(\Rightarrow19x-12=24\)
\(\Rightarrow19x=36\)
\(\Rightarrow x=\dfrac{36}{19}\)
Vậy...
\(b,6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)
\(\Rightarrow6x^2-\left(6x^2+4x-9x-6\right)-1=0\)
\(\Rightarrow6x^2-6x^2-4x+9x+6-1=0\)
\(\Rightarrow5x+5=0\)
\(\Rightarrow5x=-5\)
\(\Rightarrow x=-1\)
Vậy...
\(c,4x^2-12x=9\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+3^2-9=9\)
\(\Rightarrow\left(2x-3\right)^2-9=9\)
\(\Rightarrow\left(2x-3\right)^2=18\)
\(\Rightarrow2x-3=\sqrt{18}\)
\(\Rightarrow2x=\sqrt{18}+3\)
\(\Rightarrow x=\dfrac{\sqrt{18}+3}{2}\)
Vậy...
Chúc bạn hok tốt!!!
