Jawapan:
Penjelasan:
# "ambil perhatian bahawa" sqrtaxxsqrta = (sqrta) ^ 2 = a #
# y = sqrt ((4x + 1) / (3x-3)) #
#color (biru) "menjaringkan kedua-dua belah" #
# y ^ 2 = (sqrt ((4x + 1) / (3x-3))) ^ 2 #
# rArry ^ 2 = (4x + 1) / (3x-3) #
# rArry ^ 2 (3x-3) = 4x + 1larrcolor (biru) "cross-multiplying" #
# rArr3xy ^ 2-3y ^ 2 = 4x + 1 #
# rArr3xy ^ 2-4x = 1 + 3y ^ 2larrcolor (biru) "mengumpul istilah dalam x" #
#rArrx (3y ^ 2-4) = 1 + 3y ^ 2larrcolor (blue) "factorising" #
# rArrx = (1 + 3y ^ 2) / (3y ^ 2-4) hingga (y! = + - 4/3) #
#color (biru) "Sebagai cek" #
# "mari x = 2" #
# "maka" y = sqrt (9/3) = sqrt3 #
# "ganti ke dalam ungkapan untuk x kita harus mendapat 2" #
# x = (1 + 3 (sqrt3) ^ 2) / (3 (sqrt (3) ^ 2-4)) = (1 + 9) / (9-4) = 10 /
Bagaimana anda mempermudahkan (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1) (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), a> 1?
Memformat matematik besar ...> warna (biru) ((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1) (a + 1) - (a + 1) sqrt (a-1))) = warna (merah) (((1 / sqrt (a- 1) + sqrt (a + 1)) / (sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a-1) Cdot sqrt (a-1) cdot sqrt (a-1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) biru) (/ (1 / sqrt (a-1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt -1)))) / (sqrt (a + 1) / (sqrt (a + 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + (1 / sqrt (a-1) + sqrt (a + 1)) / sqrt (a-1) -sqrt (a + 1) sqrt (a-1) (sqrt (a-1) -sqrt (a + 1))) / sqrt (a + 1) =
Selesaikan ... 5 - x = sqrt (x + sqrt (x + sqrt (x + sqrtx))) Cari x?
Jawapannya ialah = 5-sqrt5 Let y = sqrt (x + sqrt (x + sqrt (x + sqrt (x + Oleh itu, y = 5-xy ^ 2 = x + 5-x = 5 y = + - sqrt5 Oleh itu, y ^ 2 = x + y 5 = x + sqrt5 x = 5-sqrt5
Selesaikan sistem persamaan berikut: [((1), sqrt (2) x + sqrt (3) y = 0), (2), x + y = sqrt (3) -sqrt (2))]?
{(x = (3sqrt (2) -2sqrt (3)) / (sqrt (6) -2)), (y = (sqrt (6) -2) / (sqrt (2) -sqrt (3) (1) kita mempunyai sqrt (2) x + sqrt (3) y = 0 Membahagikan kedua belah pihak dengan sqrt (2) memberi kita x + sqrt (3) / sqrt (2) Jika kita menolak "(*)" daripada (2) kita memperoleh x + y- (x + sqrt (3) / sqrt (2) y) = sqrt (3) -sqrt (2) Y = sqrt (3) -sqrt (2) => y = (sqrt (3) -sqrt (2)) / (1-sqrt (3) / sqrt (2) (sqrt (6) -2) / (sqrt (2) -sqrt (3)) Jika kita menggantikan nilai yang kita dapati untuk y kembali ke "(*)" kita dapat x + sqrt (3) / sqrt (2) (sqrt (6) -2) / (sqrt (2) -sqrt (3)) = 0 => x + (3sqrt (2