Question

[Le Trang Nhung]

Giải phương trình:

1)\(\sqrt{x}+\sqrt{x+1}=1\)

2) \(\sqrt{x+2\sqrt{2x-5}-2}+\sqrt{x-3\sqrt{2x-5}+2}=2\sqrt{x}\)

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

4) \(4\sqrt{x+1}=x^2-5x+14\)

5) \(\sqrt{x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)

Answer 1

[Trần Thành Phát Nguyễn]

can x = 1 - can x+1

bphuong

x^2=2+2 can x+1

Answer 2

[Le Trang Nhung]

1)\(\left(DKXD:x\ge0\right)\)

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

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

\(\Leftrightarrow x\left(x+1\right)=1-2x+x^2\left(0\le x\le1\right)\)

\(\Leftrightarrow x^2+x=1-2x+x^2\)

\(\Leftrightarrow3x-1=0\)

\(\Leftrightarrow x=\frac{1}{3}\)

Vậy pt có nghiệm \(x=\frac{1}{3}\)