Question

giải phương trình \(4x^2+3x+3=4\sqrt{x^3+3x^2}+2\sqrt{2x-1}\)

Answer 1

Ủa ông lớp 9 rồi ak???

Sao hôm trước bảo lp 8!!

Haizz

Answer 2

Đk \(x\ge\frac{1}{2}\)

Pt \(\Leftrightarrow4x^2+3x-7=4\left(\sqrt{x^3+3x^2}-2\right)+2\left(\sqrt{2x-1-1}\right)\)

\(\Leftrightarrow4\frac{\left(x-1\right)\left(x+2\right)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-\left(x-1\right)\left(4x+7=0\right)\)

\(\Leftrightarrow\left(x-1\right)[\frac{4\left(x+2^2\right)}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-\left(4x+7\right)=0\)

\(\Leftrightarrow x=1\)Và \(\frac{4\left(x+2\right)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0\)( *)

Xét hàm số \(f\left(x\right)=\frac{4\left(x+2\right)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\)\(\in[\frac{1}{2};+\infty]\)

Thì \(f\left(x\right)>0,\forall x\in[\frac{1}{2};+\infty]\)

=> Phương trình ( *) vô nghiệm.