Đáp án:
Giải thích các bước giải:
Bài 1:
a) `A=5.\sqrt{\frac{1}{5}}+1/2\sqrt{20}-\sqrt{45}`
`A=\sqrt{5^{2}.\frac{1}{5}}+1/2.\sqrt{4}.\sqrt{5}-\sqrt{9}.\sqrt{5}`
`A=\sqrt{5}+1/2 . 2 . \sqrt{5}-3\sqrt{5}`
`A=\sqrt{5}+\sqrt{5}-3\sqrt{5}`
`A=-\sqrt{5}`
b) `B=(\frac{5-\sqrt{5}}{\sqrt{5}}-5)(\frac{5+\sqrt{5}}{1+\sqrt{5}}+6)`
`B=[\frac{\sqrt{5}(\sqrt{5}-1)}{\sqrt{5}}-5][\frac{\sqrt{5}(\sqrt{5}+1)}{1+\sqrt{5}}+6]`
`B=(\sqrt{5}-1-5)(\sqrt{5}+6)`
`B=(\sqrt{5}-6)(\sqrt{5}+6)`
`B=-31`
2)
a) `\sqrt{4x+20}-3\sqrt{x+5}+\sqrt{16x+80}=15`
ĐK: `x \ge -5`
`⇔ \sqrt{4(x+5)}-3\sqrt{x+5}+\sqrt{16(x+5)}=15`
`⇔ \sqrt{4(x+5)}-3\sqrt{x+5}+\sqrt{16(x+5)}=15`
`⇔ 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=15`
`⇔ 3\sqrt{x+5}=15`
`⇔ \sqrt{x+5}=5`
`⇔ x+5=25`
`⇔ x=20\ (TM)`
Vậy `S={20}`
b) `x-\sqrt{x-1}=3`
`⇔ x-3=\sqrt{x-1}`
DK: \(\begin{cases} x-3 \ge 0\\x-1 \ge 0\end{cases}\)
`⇔` \(\begin{cases} x \ge 3\\x \ge 1\end{cases}\)
`⇒ x \ge 3`
`⇔ x^2-6x+9=x-1`
`⇔ x^2-7x+10=0`
`⇔ (x-5)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x=2\ (L)\\x=5\ (TM)\end{array} \right.\)
Vậy `S={5}`
