Question

[Nguyễn Bá Minh]

Giải phương trình :a,\(\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\)

b,\(\sqrt{x^2 +1-2x}+\sqrt{x^2+4-4x}=\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}\)

c,\(x^2-x-1=\sqrt{8x+1}\)

Answer 1

[Nguyễn Thiều Công Thành]

a,

\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)

\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)

b,tự nàm

c,

\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)

\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)

đặt \(\sqrt{8x+1}=a\)

=>a⁴=10a²+64a+55

nhận thấy phương trình có dạng x⁴=ax²+bx+c

tìm số m sao cho b²-4(2m+a)(m²+c)=0

sau đó đưa về (x²+m)²=k² với k là 1 số bất kì,sau đó giải ra

Answer 2

[Phan Văn Hiếu]

b)đk \(x\ge1\)

 \(\sqrt{1+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}=\sqrt{\frac{\left(x+1\right)^2+x^2.\left(x+1\right)^2+x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)

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

\(=\sqrt{\frac{\left(x^2+x+1\right)^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)

\(=\frac{x^2+x+1}{x+1}+\frac{x}{x+1}=x+1\)

\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)

\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)

\(\Leftrightarrow x+\left|x-2\right|=2014\)

giai 2 pt 

pt1 x+x-2=2014

x=1008

pt2 x+2-x=2014(vô lý)