Đáp án:
Giải thích các bước giải:
a) `\sqrt{2x-1}=\sqrt{2}-1`
ĐK: `x \ge 1/2`
`⇔ 2x-1=(\sqrt{2}-1)^2`
`⇔ 2x-1=3-2\sqrt{2}`
`⇔ 2x=4-2\sqrt{2}`
`⇔ x=2-\sqrt{2}`
Vậy `S={2-\sqrt{2}}`
b) `\sqrt{x+5}=\sqrt{7}-3`
ĐK: `x \ge -5`
`⇔ x+5=(\sqrt{7}-3)^2`
`⇔ x+5=16-6\sqrt{7}`
`⇔ x=11-6\sqrt{7}`
Vậy `S={11-6\sqrt{7}}`
c) `\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}`
ĐK: `x^2-6x+9 \ge 0` (lđ với mọi x)
`⇔ x^2-6x+9=4+2\sqrt{3}`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{6+\sqrt{16+8\sqrt{3}}}{2}\\x=\dfrac{6-\sqrt{16+8\sqrt{3}}}{2}\end{array} \right.\)
Vậy `S={\frac{6+\sqrt{16+8\sqrt{3}}}{2};\frac{6-\sqrt{16+8\sqrt{3}}}{2}}`
d) `\sqrt{3x^2-4x}=2x-3`
ĐK: \(\begin{cases} 3x^2-4x \ge 0\\2x-3 \ge 0\end{cases}\)
`⇔` \(\begin{cases} \left[ \begin{array}{l}x \le 0\\x \ge \dfrac{4}{3}\end{array} \right.\\x \ge \dfrac{3}{2}\end{cases}\)
`⇒ x \ge 3/2`
`⇔ 3x^2-4x=4x^2-12x+9`
`⇔ x^2-8x+9=0`
`⇔` \(\left[ \begin{array}{l}x=4+\sqrt{7}\ (TM)\\x=4-\sqrt{7}\ (L)\end{array} \right.\)
vậy `S={4+\sqrt{7}}`
