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: \(\sqrt{x+1}+\sqrt{9x+9}=4\)
\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)
\(\Leftrightarrow4\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Rightarrow x=0\)
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
Xin lỗi mk ghi nhầm phần
Phần 6 ban nãy là phần 5 và cho mk sửa lại
5) đk: \(x\ge\frac{3}{2}\)
Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\left(ktm\right)\)
Vậy PT vô nghiệm
6) đk: \(x\ge-\frac{7}{5}\)
Ta có: \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)
\(\Leftrightarrow\frac{5x+7}{x+3}=16\)
\(\Leftrightarrow16x+48=5x+7\)
\(\Leftrightarrow11x=-41\)
\(\Rightarrow x=-\frac{41}{11}\)(ktm)
Vậy PT vô nghiệm
7) đk: \(x\ge1\)
Ta có: \(\sqrt{5x-5}-\sqrt{35}=0\)
\(\Leftrightarrow\sqrt{5x-5}=\sqrt{35}\)
\(\Leftrightarrow5x-5=35\)
\(\Leftrightarrow5x=40\)
\(\Rightarrow x=8\left(tm\right)\)
Vậy x = 8
1. \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)
<=> \(\sqrt{x-1}=\sqrt{2x\left(x-1\right)}\)
ĐK : x ≥ 1
<=> \(x-1=2x\left(x-1\right)\)
<=> \(2x^2-2x-x+1=0\)
<=> \(2x\left(x-1\right)-\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=\frac{1}{2}\left(ktm\right)\end{cases}}\)
\(\sqrt{5x^2}=2x-1\)
ĐK : x ≥ 1/2
<=> \(5x^2=4x^2-4x+1\)
<=> \(5x^2-4x^2+4x-1=0\)
<=> \(x^2+4x-1=0\)
<=> \(\left(x^2+4x+4\right)-5=0\)
<=> \(\left(x+2\right)^2-\left(\sqrt{5}\right)^2=0\)
<=> \(\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)
<=> \(\orbr{\begin{cases}x+2-\sqrt{5}=0\\x+2+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}-2\\x=-\sqrt{5}-2\end{cases}\left(ktm\right)}\)
3. \(\sqrt{x+1}+\sqrt{9x+9}=4\)
ĐK : x ≥ -1
<=> \(\sqrt{x+1}+\sqrt{3^2\left(x+1\right)}=4\)
<=> \(\sqrt{x+1}+3\sqrt{x+1}=4\)
<=> \(\sqrt{x+1}\cdot\left(1+3\right)=4\)
<=> \(\sqrt{x+1}\cdot4=4\)
<=> \(\sqrt{x+1}=1\)
<=> \(x+1=1\)
<=> \(x=0\left(tm\right)\)
4. \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)
<=> \(\sqrt{x-2}=\sqrt{x\left(x-2\right)}\)
ĐK : x ≥ 2
<=> \(x-2=x\left(x-2\right)\)
<=> \(x\left(x-2\right)-x+2=0\)
<=> \(x\left(x-2\right)-\left(x-2\right)=0\)
<=> \(\left(x-2\right)\left(x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)
5. \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
ĐK : x ≥ 3/2
<=> \(\frac{2x-3}{x-1}=4\)
<=> 2x - 3 = 4( x - 1 )
<=> 2x - 3 = 4x - 4
<=> -3 + 4 = 4x - 2x
<=> 1 = 2x
<=> x = 1/2 ( ktm )
6. \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)
ĐK : x ≥ -7/5
<=> \(\frac{5x+7}{x+3}=16\)
<=> 5x + 7 = 16( x + 3 )
<=> 5x + 7 = 16x + 48
<=> 7 - 48 = 16x - 5x
<=> -41 = 11x
<=> x = -41/11 ( ktm )
7. \(\sqrt{5x-5}-\sqrt{35}=0\)
<=> \(\sqrt{5x-5}=\sqrt{35}\)
ĐK : x ≥ 1
<=> 5x - 5 = 35
<=> 5x = 40
<=> x = 8 ( tm )
