Được câu d rồi mà không biết có đúng không.
\(d)\left(x+3\right)\left(x+2\right)+3x=4\left(x+\dfrac{3}{4}\right)\)
\(x^2+2x+3x+6+3x=4x+3\)
\(x^2+8x+6-4x=3\)
\(x^2+4x+4-4+6=3\)
\(\left(x+2\right)^2+2=3\)
\(\left(x+2\right)^2=3-2=1^2\)
Ta có:
Vậy: \(x=-1\) hoặc \(x=-3\)
\(a)\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(3x^3-4x^2+2x-1-3x^3+4x^2\) \(=\dfrac{5}{2}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{5}{2}+1\) \(=\dfrac{7}{2}\)
\(x=\dfrac{7}{2}:2=\dfrac{7}{2}\cdot\dfrac{1}{2}\)
Vậy: \(x=\dfrac{7}{4}\)
\(b)2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(2x^2+3x^2-3=5x^2+5x\)
\(5x^2-3=5x^2+5x\)
\(-3=5x^2-5x^2+5x\)
\(-5x=3\)
Vậy: \(x=-\dfrac{3}{5}\)
\(c)\left(x^2-x+1\right)\left(x+1\right)-\left(x^3-3x\right)=15\)
\(x^3+1-x^3+3x=15\)
\(3x+1=15\)
\(3x=15-1=14\)
Vậy: \(x=\dfrac{14}{3}\)
