m)
\(\sqrt{x+1}-\sqrt{x-2}=1\) (ĐK: x ≥ 2)
=> \(\sqrt{x+1}=1+\sqrt{x-2}\)
=> \(x+1=1+x-2+2\sqrt{x-2}\)
=> \(-2+2\sqrt{x-2}=0\)
\(\Rightarrow1-\sqrt{x-2}=0\)
=> x - 2 = 1
=> x = 3
n) \(\sqrt{x-5}-\sqrt{x+4}=2\) (ĐK: x ≥ 5)
=> \(\sqrt{x-5}=2+\sqrt{x+4}\)
\(\Rightarrow x-5=2+x+4+4\sqrt{x+4}\)
\(\Rightarrow-11=4\sqrt{x+4}\)
Mà: \(4\sqrt{x+4}\ge0\)
=> PT vô nghiệm
o) Làm tương tự 2 câu trên nhé!
p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (ĐK: x ≥ 2)
\(\Rightarrow2x-1+x-2+2\sqrt{\left(2x-1\right)\left(x-2\right)}=x+1\)
\(\Rightarrow2x-4+2\sqrt{\left(2x-1\right)\left(x-2\right)}=0\)
*Với ĐK: x ≥ 2 thì 2x - 4 ≥ 0
Lại có: \(2\sqrt{\left(2x-1\right)\left(x-2\right)}\ge0\)
Mà: \(2x-4+2\sqrt{\left(2x-1\right)\left(x-2\right)}=0\)
=> x = 2 (t/m đk)
