Đáp án:
Giải thích các bước giải:
a) `\sqrt{3}x-\sqrt{75}=0`
`⇔ \sqrt{3}x=\sqrt{75}`
`⇔ x=\frac{\sqrt{75}}{\sqrt{3}}`
`⇔ x=\sqrt{\frac{75}{3}}`
`⇔ x=\sqrt{25}=5`
Vậy `S={5}`
b) `\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}`
`⇔ \sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}`
`⇔ \sqrt{2}x=4\sqrt{2}`
`⇔ x=\frac{4\sqrt{2}}{\sqrt{2}}`
`⇔ x=4`
Vậy `S={4}`
c) `\sqrt{10(x-3)}=\sqrt{26}`
ĐK: `x \ge 3`
`⇔ 10(x-3)=26`
`⇔ 10x-30=26`
`⇔ 10x=56`
`⇔ x=56/10\ (TM)`
Vậy `S={56/10}`
d) `\sqrt{3x^2}=x+2`
ĐK: `x \ge -2` (do `3x^2 ≥0 ∀x)`
`⇔ 3x^2=x^2+4x+4`
`⇔ 2x^2-4x-4=0`
`⇔ x^2-2x-2=0`
`Δ'=(-1)^2-1.(-2)=3`
`Δ'>0:` PT có 2 nghiệm pb
`x_{1}=1+\sqrt{3}\ (TM),x_{2}=1-\sqrt{3}\ (TM)`
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
e) `1+\sqrt{3x+1}=3x`
`⇔ \sqrt{3x+1}=3x-1`
ĐK: \(\begin{cases} 3x+1 \ge 0\\3x-1 \ge 0\end{cases}\)
`⇔` \(\begin{cases} x \ge -\dfrac{1}{3}\\x \ge \dfrac{1}{3}\end{cases}\)
`⇒ x \ge 1/3`
`⇔ 3x+1=9x^2-6x+1`
`⇔ 9x^2-9x=0`
`⇔ 9x(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x=0\ (L)\\x=1\ (TM)\end{array} \right.\)
Vậy `S={1}`
f) `\sqrt{x^2+6x+9}=3x-6`
ĐK: `x \ge 2` (do `x^2+6x+9 \ge 0∀x)`
`⇔ x^2+6x+9=9x^2-36x+36`
`⇔ 8x^2-42x+27=0`
`⇔ (2x-9)(4x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{9}{2}\ (TM)\\x=\dfrac{3}{4}\ (L)\end{array} \right.\)
Vậy `S={9/2}`
g) `\sqrt{x^2-4x+4}-2x+5=0`
`⇔ \sqrt{x^2-4x+4}=2x-5`
ĐK: `x \ge 5/2`
`⇔ x^2-4x+4=4x^2-20x+25`
`⇔ 3x^2-16x+21=0`
`⇔ (x-3)(3x-7)=0`
`⇔` \(\left[ \begin{array}{l}x=3\ (TM)\\x=\dfrac{7}{3}\ (L)\end{array} \right.\)
Vậy `S={3}`
h) `\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}`
ĐK: \(\begin{cases} 3x-5 \ge 0\\x+1 \ge 0\end{cases}\)
`⇔` \(\begin{cases} x \ge \dfrac{5}{3}\\x \ge -1\end{cases}\)
`⇒ x \ge 5/3`
`⇔ 2+\sqrt{3x-5}=x+1`
`⇔ \sqrt{3x-5}=x-1` (ĐK : `x \ge 5/3)`
`⇔ 3x-5=x^2-2x+1`
`⇔ x^2-5x+6=0`
`⇔ (x-2)(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) (TM)
Vậy `S={2;3}`
i) `\sqrt{\frac{3x-1}{x+2}}=\sqrt{5}`
ĐK:
TH1: \(\begin{cases} 3x-1 \ge 0\\x+2 > 0\end{cases}\)
`⇔` \(\begin{cases} x \ge \dfrac{1}{3}\\x > -2\end{cases}\)
`⇒ x \ge 1/3`
TH2: \(\begin{cases} 3x-1 \le 0\\x+2 < 0\end{cases}\)
`⇔` \(\begin{cases} x \le \dfrac{1}{3}\\x < -2\end{cases}\)
`⇒ x < -2`
`⇔ \frac{\sqrt{3x-1}}{\sqrt{x+2}}=\sqrt{5}`
`⇔ \sqrt{5}(\sqrt{x+2})=\sqrt{3x-1}`
`⇔ 5(x+2)=3x-1`
`⇔ 5x+10=3x-1`
`⇔ 2x=-11`
`⇔ x=-11/2\ (TM)`
Vậy `S={-11/2}`
k) `\frac{x^2}{\sqrt{7}}-\sqrt{28}=0`
`⇔ \frac{x^2}{\sqrt{7}}=\sqrt{28}`
`⇔ x^2=\sqrt{28.7}`
`⇔ x^2=14`
`⇔ x=±\sqrt{14}`
Vậy `S={-\sqrt{14},\sqrt{14}}`
m) `\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4`
`⇔ 4(\sqrt{x+3})=\sqrt{5x+7}`
ĐK: `x \ge -7/5`
`⇔ 16(x+3)=5x+7`
`⇔ 16x+48=5x+7`
`⇔ 11x=-41`
`⇔ x=-41/11\ (L)`
Vậy PT vô nghiệm
