Đáp án:
Bài 1:
`a, 3√3+4√12-5√27`
`= 3√3+4`$\sqrt[]{4.3}$ -5$\sqrt[]{9.3}$
`= 3√3+4.√4.√3-5.√9.√3`
`= 3√3+4.2√3-5.3√3`
`= 3√3+8√3-15√3`
`= (3+8-15)√3`
`= -4√3`
`b, 1/2 √48-2√75-(√33)/(√11)+5√1 1/3`
`= 1/2`$\sqrt[]{16.3}$ `- 2`$\sqrt[]{25.3}$ `-√33/11+5√4/3`
`= 1/2. 4√3 - 2. 5√3 -√3+ 5. 2/(√3)`
`= 2√3-10√3-√3+10/(√3)`
`= (2-10-3)√3+10/(√3)`
`= -11√3+10/(√3)`
`= ((-11).√3. √3)/(√3)+10/(√3)`
`= (-33)/(√3)+10/(√3)`
`= (-23)/(√3)`
`c, A =` $\sqrt[]{(1-√3)^2}$-$\sqrt[]{(√3+2)^2}$
`= |1-√3|-|√3+2|`
`= √3-1-√3-2`
`= -3`
`d, B = `$\sqrt[]{(2-√3)^2}$-$\sqrt[]{4-2√3}$
`= |2-√3|-`$\sqrt[]{3-2√3+1}$
`= 2-√3-`$\sqrt[]{(√3-1)^2}$
`= 2-√3-|√3-1|`
`= 2-√3-√3+1`
`= 3-2√3`
Bài 2:
a, $\sqrt[]{1-4x+4x^2}$ `= 5 (ĐKXĐ : x≤1/2)`
⇔ $\sqrt[]{(1-2x)^2}$ `= 5`
`⇔ |1-2x| = 5`
`⇒ TH1 : 1-2x = 5`
`⇔ 2x = -4`
`⇔ x = -2` (thỏa mãn)
`TH2 : 1-2x = -5`
`⇔ 2x = 6`
`⇔ x = 3` (không thỏa mãn)
+Vậy `S = {-2}`
b, $\sqrt[]{4-5x}$ `= 12 (ĐKXĐ : x≤4/5)`
`⇔ 4-5x = 144`
`⇔ 5x = -100`
`⇔ x = -24` (thỏa mãn)
+Vậy `S = {-24}`
c, $\sqrt[]{4x+20}$`-3`$\sqrt[]{5+x}$`+4/3` $\sqrt[]{9x+45}$ `= 6 (ĐKXĐ : x≥-5)`
⇔ $\sqrt[]{4(x+5)}$`-3`$\sqrt[]{x+5}$`+4/3` $\sqrt[]{9(x+5)}$ `= 6`
`⇔ 2`$\sqrt[]{x+5}$`-3`$\sqrt[]{x+5}$`+4/3. 3`$\sqrt[]{x+5}$ `= 6`
`⇔ 2`$\sqrt[]{x+5}$`-3`$\sqrt[]{x+5}$`+4`$\sqrt[]{x+5}$ `= 6`
`⇔ (2-3+4)`$\sqrt[]{x+5}$ `= 6`
`⇔ 3`$\sqrt[]{x+5}$ `= 6`
⇔ $\sqrt[]{x+5}$ `= 2`
`⇔ x+5 = 4`
`⇔ x = -1` (thỏa mãn)
+Vậy `S = {-1}`
