Bài 11:
a: Ta có: \(\left(x^2+2x+4\right)\left(2-x\right)+x\left(x-3\right)\left(x+4\right)-x^2+24=0\)
\(\Leftrightarrow8-x^3-x^2+24+x\left(x^2+x-12\right)=0\)
\(\Leftrightarrow-x^3-x^2+32+x^3+x^2-12x=0\)
\(\Leftrightarrow12x=32\)
hay \(x=\dfrac{8}{3}\)
b: Ta có: \(\left(\dfrac{1}{2}x+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\dfrac{1}{4}x+3\right)=0\)
\(\Leftrightarrow\dfrac{5}{2}x-3x^2+15-18x+3x^2+36x-\dfrac{1}{2}x-\dfrac{3}{2}=0\)
\(\Leftrightarrow x\cdot20=-\dfrac{27}{2}\)
hay \(x=-\dfrac{27}{40}\)
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