Question

[Đào Linh Chi]

giải pt \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)

Answer 1

[Incursion_03]

Lần sau đừng tự tiện xếp vào phần bất pt bạn nhé :(

Ta có : \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)

\(\Leftrightarrow4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)=3x^2\)

\(\Leftrightarrow4\left(x^2+17x+60\right)\left(x^2+16x+60\right)=3x^2\)(1)

Đặt \(x^2+16x+60=a\)

Pt (1) \(\Leftrightarrow4\left(a+x\right)a=3x^2\)

         \(\Leftrightarrow4\left(a^2+ax\right)=3x^2\)

          \(\Leftrightarrow4a^2+4ax=3x^2\)

          \(\Leftrightarrow4a^2+4ax+x^2=4x^2\)

         \(\Leftrightarrow\left(2a+x\right)^2=4x^2\)

          \(\Leftrightarrow\orbr{\begin{cases}2a+x=2x\\2a+x=-2x\end{cases}}\)

*Nếu \(2a+x=2x\)

\(\Leftrightarrow2a=x\)

\(\Leftrightarrow x^2+16x+60=x\)

\(\Leftrightarrow x^2+15x+60=0\)

\(\Leftrightarrow x^2+2.\frac{15}{2}.x+\frac{225}{4}+\frac{15}{4}=0\)

\(\Leftrightarrow\left(x+\frac{15}{2}\right)^2+\frac{15}{4}=0\)

Pt vô nghiệm

*Nếu \(2a+x=-2x\)

\(\Leftrightarrow2a+3x=0\)

\(\Leftrightarrow2\left(x^2-16x+60\right)+3x=0\)

\(\Leftrightarrow2x^2-32x+120+3x=0\)

\(\Leftrightarrow2x^2-29x+120=0\)

\(\Leftrightarrow x^2-\frac{29}{2}x+60=0\)

\(\Leftrightarrow x^2-2.\frac{29}{4}.x+\frac{841}{16}+\frac{119}{16}=0\)

\(\Leftrightarrow\left(x-\frac{29}{4}\right)^2+\frac{119}{16}=0\)

Pt vô nghiệm

Vậy pt vô nghiệm