`a.sqrt{2x-5}=7`
Điều kiện:`2x-5>=0=>x>=5/2`
`⇔|2x-5|=49`
`⇔`\(\left[ \begin{array}{l}2x-5=49\\2x-5=-49\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=54\\2x=-44\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=27\text{(nhận)}\\x=-22\text{(loại)}\end{array} \right.\)
Vậy `S={27}`
`b.sqrt{x+3}+2sqrt{9x+27}=6`
Điều kiện:`x+3>=0=>x>=-3`
`⇔sqrt{x+3}+2.sqrt{x+3}=6`
`⇔(sqrt{x+3})(1+6)=6`
`⇔sqrt{x+3}=6/7`
`⇔|x+3|=36/49`
`⇔`\(\left[ \begin{array}{l}x+3=\dfrac{36}{49}\\x+3=-\dfrac{36}{49}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{119}{49}\text{(nhận)}\\x=-\dfrac{183}{49}\text{(loại)}\end{array} \right.\)
Vậy `S={-119/49}`
`c.sqrt{4.(x^2+6x+9)}=14`
Điều kiện:`4.(x^2+6x+9)>=0=>4.(x+3)^2>=0`(đúng)
`⇔2sqrt{(x+3)^2}=14`
`⇔2|x+3|=14`
`⇔`\(\left[ \begin{array}{l}2(x+3)=14\\2(x+3)=-14\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x+6=14\\2x+6=-14\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=8\\2x=-20\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=4\\x=-10\end{array} \right.\)
Vậy `S={4;=10}.`
