\(\left(x+1\right).\left(x-2\right)-\left(4x+1\right).\left(x+2\right)\)
\(\Leftrightarrow x^2-x-2-\left(4x^2+9x+2\right)\)
\(\Leftrightarrow x^2-x-2-4x^2-9x-2\)
\(\Leftrightarrow-3x^2-10x-4\)
hơi sai sai
ko có dấu bằng thì tìm x bằng niềm tin
`4. ( x+1).( x-2) - ( 4x+1).(x+2)=66`
`4x^2 -4x-8 - 4x^2 -9x-2=66`
`-13x-10=66`
`-13x = 66 + 10`
`-13x=76`
`x=76:(-13)`
`x=(-76)/13`
\(\left(x+1\right).\left(x-2\right)-\left(4x+1\right).\left(x+2\right)=66\)
\(\Leftrightarrow x^2-2x+x-2-\left(4x^2+x+8x+2\right)=66\)
\(\Leftrightarrow x^2-x-2-4x^2-x-8x-2-66=0\)
\(\Leftrightarrow3x^2+10x+70=0\)
\(\Leftrightarrow3x^2+2.\left(\sqrt{3}\right).\left(\dfrac{5\sqrt{3}}{3}\right)x+\dfrac{25}{3}+\dfrac{182}{3}=0\)
\(\Leftrightarrow\left(\sqrt{3}x+\dfrac{5\sqrt{3}}{3}\right)^2+\dfrac{182}{3}=0\)
\(\Leftrightarrow\left(\sqrt{3}x+\dfrac{5\sqrt{3}}{3}\right)^2=-\dfrac{182}{3}\)(Vô lí)
Vậy phương trình vô nghiệm
