i. sqrt(3) * (sqrt(2) - sqrt(3)) ^ 2 - (sqrt(3) + sqrt(2))
k. (1 - sqrt(3)) ^ 2 * (1 + 2sqrt(3)) ^ 2
n(1/(sqrt(5) - sqrt(2)) - 1/(sqrt(5) + sqrt(2)) + 1) * 1/(

Question

i. sqrt(3) * (sqrt(2) - sqrt(3)) ^ 2 - (sqrt(3) + sqrt(2))
k. (1 - sqrt(3)) ^ 2 * (1 + 2sqrt(3)) ^ 2
n(1/(sqrt(5) - sqrt(2)) - 1/(sqrt(5) + sqrt(2)) + 1) * 1/((sqrt(2) + 1) ^ 2)

Pikasonic · Answer

`i. sqrt(3) * (sqrt(2) - sqrt(3)) ^ 2 - (sqrt(3) + sqrt(2))`
`= \sqrt3 . [(\sqrt2)^2 - 2.\sqrt2 . \sqrt3 + (\sqrt3)^2]  - \sqrt3 - \sqrt2`
`= \sqrt3 . [2 - 2.\sqrt6  + 3]  - \sqrt3 - \sqrt2`
`= \sqrt3 . [5 - 2.\sqrt6 ]  - \sqrt3 - \sqrt2`
`= 5\sqrt3 - 2\sqrt(18) - \sqrt3 -\sqrt2`
`= 5\sqrt3 - 2\sqrt(9.2) - \sqrt3 -\sqrt2`
`= 5\sqrt3 - 2.3.\sqrt(2) - \sqrt3 -\sqrt2`
`= 5\sqrt3 - 6\sqrt(2) - \sqrt3 -\sqrt2`
`= 4\sqrt3 - 7\sqrt2`
`--------------------`
`k. (1 - sqrt(3)) ^ 2 * (1 + 2sqrt(3)) ^ 2`
`= [(1-\sqrt3)(1+2\sqrt3)]^2`
`= [1 + 2\sqrt3 - \sqrt3 - 2(\sqrt3)^2]^2`
`= (1 + 2\sqrt3 - \sqrt3 - 2.3)^2`
`= (1 + 2\sqrt3 - \sqrt3 - 6)^2`
`= ( \sqrt3 - 5 )^2`
`= (\sqrt3)^2 - 2.\sqrt3 .5 + 5^2`
`= 3 - 10\sqrt3 + 25`
`= 28 - 10\sqrt3`
`------------------`
`n. (1/(sqrt(5) - sqrt(2)) - 1/(sqrt(5) + sqrt(2)) + 1) * 1/((sqrt(2) + 1) ^ 2)`
`= [(sqrt(5) + sqrt(2))/((sqrt(5) - sqrt(2))(sqrt(5) + sqrt(2)) ) - (sqrt(5) - sqrt(2))/((sqrt(5) + sqrt(2))(sqrt(5) + sqrt(2))) + 1] * 1/((sqrt(2) + 1)^2`
`= [(sqrt(5) + sqrt(2))/((sqrt(5))^2 - (sqrt(2))^2 ) - (sqrt(5) - sqrt(2))/((sqrt(5))^2 - (sqrt(2))^2) + 1] * 1/((sqrt(2) + 1)^2`
`= [(sqrt(5) + sqrt(2))/(3 ) - (sqrt(5) - sqrt(2))/(3) + 3/3] * 1/((sqrt(2) + 1)^2`
`= (sqrt(5) + sqrt(2)- (sqrt(5) - sqrt(2)) + 3)/(3 )  * 1/((sqrt(2))^2 + 2\sqrt2 + 1)`
 `= (sqrt(5) + sqrt(2)- sqrt(5) + sqrt(2) + 3)/(3 )  * 1/(2+ 2sqrt(2) + 1)`
 `= ( 2sqrt(2) + 3)/(3 )  * 1/((2sqrt(2) +3)`
`= ( 2sqrt(2) + 3)/(3(2sqrt(2) + 3) ) `
`=1/3`

hieunguyen529 · Answer

`# Ly`
`i)`
     `sqrt3 . (sqrt2 -sqrt3)^2 - (sqrt3 +sqrt2)`
`= sqrt3 . [(sqrt2)^2 - 2.sqrt2 . sqrt3 + (sqrt3)^2] - sqrt3 - sqrt2`
`= sqrt3 . (2 - 2sqrt6 + 3) - sqrt3 - sqrt2`
`= sqrt3 . (5- 2sqrt6) - sqrt3 - sqrt2`
`= 5sqrt3 - 2sqrt{18} - sqrt3 - sqrt2`
`= 5sqrt3 - 2. sqrt{3^2 . 2} - sqrt3 - sqrt2`
`= 5sqrt3 - 2. 3sqrt2 - sqrt3 - sqrt2`
`= 5sqrt3 - 6sqrt2 - sqrt3 - sqrt2`
`= (5sqrt3 - sqrt3) - (6sqrt2 + sqrt2)`
`= 4sqrt3 - 7sqrt2`
`k)`
                `(1- sqrt3)^2 . (1+ 2sqrt3)^2`
`= [1^2 - 2.1.sqrt3 + (sqrt3)^2] . [1^2 + 2.1.2sqrt3 + (2sqrt3)^2]`
`= (1- 2sqrt3 + 3)(1 + 4sqrt3 + 12)`
`= (4- 2sqrt3)(13 + 4sqrt3)`
`= 4 (13 + 4sqrt3) - 2sqrt3 (13 + 4sqrt3)`
`= 52 + 16sqrt3 - 26sqrt3 - 8.3`
`= 52 + 16sqrt3 - 26sqrt3 - 24`
`= (52 - 24) + (16 sqrt3 - 26sqrt3)`
`= 28 -10sqrt3`
`k)`
         `(1/(sqrt5 -sqrt2) - 1/(sqrt5 +sqrt2) +1) . 1/(sqrt2 +1)^2`     
`= [(sqrt5 +sqrt2 - (sqrt5 -sqrt2))/((sqrt5 -sqrt2)(sqrt5 +sqrt2)) +1] . 1/((sqrt2)^2 + 2.sqrt2 .1 + 1^2)`
`= [(sqrt5 +sqrt2 - sqrt5 +sqrt2)/((sqrt5)^2 - (sqrt2)^2) +1] . 1/(2 + 2sqrt2 +1)`
`= ((2sqrt2)/(5- 2) +1) . 1/(3+ 2sqrt2)`
`= ((2sqrt2)/3 + 3/3) . 1/(3+ 2sqrt2)`
`= (3+ 2sqrt2)/3 . 1/(3+ 2sqrt2)`
`= 1/3`

i. sqrt(3) * (sqrt(2) - sqrt(3)) ^ 2 - (sqrt(3) + sqrt(2)) k. (1 - sqrt(3)) ^ 2 * (1 + 2sqrt(3)) ^ 2 n(1/(sqrt(5) - sqrt(2)) - 1/(sqrt(5) + sqrt(2)) + 1) * 1/((sqrt(2) + 1) ^ 2)

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i. sqrt(3) * (sqrt(2) - sqrt(3)) ^ 2 - (sqrt(3) + sqrt(2)) k. (1 - sqrt(3)) ^ 2 * (1 + 2sqrt(3)) ^ 2 n(1/(sqrt(5) - sqrt(2)) - 1/(sqrt(5) + sqrt(2)) + 1) * 1/((sqrt(2) + 1) ^ 2)

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Tham Gia Group Dành Cho Lớp 9 - Ôn Thi Vào Lớp 10 Miễn Phí

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