Question

[vũ tiền châu]

giải các phương trình vô tỉ sau 

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

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

Answer 1

[Thắng Nguyễn]

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

ĐK:\(x\ge-\frac{1}{2}\)

\(\Leftrightarrow\left(x-1\right)\sqrt{x+1}+\sqrt{2x+1}-\sqrt{3}=\sqrt{x+2}-\sqrt{3}\)

\(\Leftrightarrow\left(x-1\right)\sqrt{x+1}+\frac{2x+1-3}{\sqrt{2x+1}+\sqrt{3}}=\frac{x+2-3}{\sqrt{x+2}+\sqrt{3}}\)

\(\Leftrightarrow\left(x-1\right)\sqrt{x+1}+\frac{2x-2}{\sqrt{2x+1}+\sqrt{3}}=\frac{x-1}{\sqrt{x+2}+\sqrt{3}}\)

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

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

Suy ra x=1

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

\(\Leftrightarrow\frac{1}{\left(x-1\right)^2}-4+\sqrt{3x+1}-\sqrt{\frac{5}{2}}=\frac{1}{x^2}-4+\sqrt{x+2}-\sqrt{\frac{5}{2}}\)

\(\Leftrightarrow\frac{4x^2-8x+3}{-x^2+2x-1}+\frac{3x+1-\frac{5}{2}}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}=\frac{-\left(4x^2-1\right)}{x^2}+\frac{x+2-\frac{5}{2}}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}\)

\(\Leftrightarrow\frac{2\left(x-\frac{1}{2}\right)\left(2x-3\right)}{-x^2+2x-1}+\frac{6\left(x-\frac{1}{2}\right)}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}+\frac{2\left(x-\frac{1}{2}\right)\left(2x+1\right)}{x^2}-\frac{x-\frac{1}{2}}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{2\left(2x-3\right)}{-x^2+2x-1}+\frac{6}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}+\frac{2\left(2x+1\right)}{x^2}-\frac{1}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}\right)=0\)

Suy ra x=1/2

Answer 2

[khánhchitt3003]

96 đặt\(\sqrt{x+7}+\sqrt{6-x}=a\)

=>\(a^2-13=2\sqrt{-x^2-x+42}\)

xong cậu thay vào pt là đc