Đáp án:
Giải thích các bước giải:
a) `\frac{1}{\sqrt{x-1}+\sqrt{x-2}}+\frac{1}{\sqrt{x-2}+\sqrt{x-3}}+....+\frac{1}{\sqrt{x-9}+\sqrt{x-10}}=1`
`ĐK: x \ge 10`
`⇔ \frac{\sqrt{x-1}-\sqrt{x-2}}{(\sqrt{x-1}+\sqrt{x-2}).(\sqrt{x-1}-\sqrt{x-2})}+\frac{\sqrt{x-2}-\sqrt{x-3}}{(\sqrt{x-2}+\sqrt{x-3}).(\sqrt{x-2}-\sqrt{x-3})}+....+\frac{\sqrt{x-9}-\sqrt{x-10}}{(\sqrt{x-9}+\sqrt{x-10}).(\sqrt{x-9}-\sqrt{x-10})}=1`
`⇔ \sqrt{x-1}-\sqrt{x-2}+\sqrt{x-2}-\sqrt{x-3}+....+\sqrt{x-9}-\sqrt{x-10}=1`
`⇔ \sqrt{x-1}-\sqrt{x-10}=1`
`⇔ x-1+x-10-2\sqrt{(x-1)(x-10)}=1`
`⇔ 2x-12=2\sqrt{(x-1)(x-10)}`
`⇔ x-6=\sqrt{(x-1)(x-10)}`
`⇔ x^2-12x+36=x^2-11x+10`
`⇔ x=26\ (TM)`
Vậy `S={26}`
d) `\sqrt{6-x}+\sqrt{2x+6}+\sqrt{6x-5}=x^2-2x-5`
`ĐK: 5/6 \le x \le 6`
`⇔ (\sqrt{6-x}-1)+(\sqrt{2x+6}-4)+(\sqrt{6x-5}-5)=x^2-2x-15`
`⇔ \frac{5-x}{\sqrt{6-x}+1)+\frac{2x-10}{\sqrt{2x+6}+4}+\frac{6x-30}{\sqrt{6x-5}+5}=(x+3)(x-5)`
`⇔ (x-5)(\frac{-1}{\sqrt{6-x}+1)+\frac{2}{\sqrt{2x+6}+4}+\frac{6}{\sqrt{6x-5}+5}-x-3)=0`
Do `\frac{-1}{\sqrt{6-x}+1)+\frac{2}{\sqrt{2x+6}+4}+\frac{6}{\sqrt{6x-5}+5}-x-3<0∀x`
`⇒ x-5=0⇔x=5`
Vậy `S={5}`
e) `\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}`
`ĐK: |x| \ge 1`
`⇔ x-\sqrt{x^2-1}=\frac{1}{4x}+\frac{3x}{2x^2+2}`
`⇔ \sqrt{x^2-1}=\frac{(x^2-1)(4x^2+1)}{4x(x^2+1)}`
`⇔ \sqrt{x^2-1}.[1-\frac{\sqrt{x^2-1}(4x^2+1)}{4x(x^2+1)}=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{x^2-1}=0\\ \dfrac{\sqrt{x^2-1}(4x^2+1)}{4x(x^2+1)}=1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=±1\\ \dfrac{\sqrt{x^2-1}(4x^2+1)}{4x(x^2+1)}=1\ (1)\end{array}\right.\)
Xét `x \le -1` thì VT của `(1)` là 1 số âm (vô lí)
Xét `x \ge 1`
`⇔ \sqrt{x^2-1}(4x^2+1)=4x(x^2+1)`
`⇔ 40x^4+23x^2+1=0`
`⇒` Vô nghiệm do luôn `>0∀x`
Vậy `S={-1;1}`
