Question

[tran phuong]

1) \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\) 0

2)\(\sqrt{5x^2}=2x-1\)

3) \(\sqrt{x+1}+\sqrt{9x+9}=4\)

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

5)\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

6) \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)

7) \(\sqrt{5x-5}-\sqrt{35}=0\)

Mình đang cần rất gấp=(

Answer 1

[Nguyễn Minh Đăng]

1) đk: \(x\ge1\)

Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)

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

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

\(\Leftrightarrow2x^2-3x+1=0\)

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

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

Vậy x = 1

2) đk: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{5x^2}=2x-1\)

\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)

\(\Leftrightarrow5x^2=4x^2-4x+1\)

\(\Leftrightarrow x^2+4x-1=0\)

\(\Leftrightarrow\left(x+2\right)^2-5=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)

=> PT vô nghiệm

3) đk: \(x\ge-1\)

Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)

\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)

\(\Leftrightarrow4\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=1\)

\(\Rightarrow x=0\)

Answer 2

[Nguyễn Minh Đăng]

4) đk: \(x\ge2\)

Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)

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

\(\Leftrightarrow x-2=x\left(x-2\right)\)

\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Vậy x = 2

6) đk: \(x\ge-\frac{7}{5}\)

Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\frac{2x-3}{x-1}=2\)

\(\Leftrightarrow2x-3=2x-2\)

\(\Leftrightarrow0x=1\) vô lý

=> PT vô nghiệm