Question

[Trịnh Trọng Khánh]

giải phương trình

a,2\(\sqrt{3x}\) +12=4x-5\(\sqrt{3}\)

b,2(\(\sqrt{5}\)-x)+2=4(\(\sqrt{5x}\)-5)

c,\(\sqrt{2}\)(x+5)-4x=5+4\(\sqrt{2}\)

d,\(\frac{7x+1}{2001}\)+1=\(\frac{2+3x}{2003}\)+\(\frac{2\left(2x+2001\right)}{2001}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\Leftrightarrow2\sqrt{3x}+12-4x+5\sqrt{3}=0\)

\(\Leftrightarrow-4x+2\sqrt{3}\cdot\sqrt{x}+12+5\sqrt{3}=0\)

Đặt \(\sqrt{x}=a\left(a>=0\right)\)

Phương trình trở thành \(-4a^2+2\sqrt{3}a+12+5\sqrt{3}=0\)

\(\Delta=\left(2\sqrt{3}\right)^2-4\cdot\left(-4\right)\cdot\left(12+5\sqrt{3}\right)\)

\(=12+16\left(12+5\sqrt{3}\right)\)

\(=12+192+80\sqrt{3}=204+80\sqrt{3}\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{-2\sqrt{3}-\sqrt{204+80\sqrt{3}}}{-8}=\dfrac{2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{8}\left(nhận\right)\\a_2=\dfrac{-2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{-8}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow a=\dfrac{2\sqrt{3}+2\sqrt{26+20\sqrt{3}}}{8}=\dfrac{\sqrt{3}+\sqrt{26+20\sqrt{3}}}{4}\)

\(\Leftrightarrow x=a^2\simeq5,66\)

c: \(\Leftrightarrow x\sqrt{2}+5\sqrt{2}-4x-5-4\sqrt{2}=0\)

\(\Leftrightarrow x\left(\sqrt{2}-4\right)+\sqrt{2}-5=0\)

\(\Leftrightarrow x=\dfrac{5-\sqrt{2}}{\sqrt{2}-4}=\dfrac{-18-\sqrt{2}}{14}\)

d: \(\Leftrightarrow\dfrac{7x+1-4x-4002}{2001}=\dfrac{3x+2}{2003}-1\)

\(\Leftrightarrow3x-4001=0\)

hay x=4001/3