Rút gon
√(16√(2+√(9+4√2)))
K
Khách
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11 tháng 8 2017
\(C=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
\(=\frac{5^{32}-1}{2}\)
B
29 tháng 7 2019
\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
Vây...
\(b,\)Ta có:\(x=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)
Thay \(x=\left(1-\sqrt{3}\right)^2\)vào A ta được:
\(A=\frac{\sqrt{\left(1-\sqrt{3}\right)^2}-2}{\sqrt{\left(1-\sqrt{3}\right)^2}+2}=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}=\frac{-\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=-\sqrt{3}\)
L
2
LV
3 tháng 5 2017
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{49.51}{50^2}\)
\(=\frac{2.3....49}{3.4....50}.\frac{4.5....51}{3.4....50}\)
\(=\frac{2}{50}.\frac{17}{1}\)
\(=\frac{17}{25}\)
3 tháng 5 2017
Ta có : \(A=\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)
\(A=\frac{8.15.....2499}{9.16.....2500}\)
\(A=\frac{\left(2.4\right).\left(3.5\right).....\left(49.51\right)}{\left(3.3\right).\left(4.4\right).....\left(50.50\right)}\)
\(A=\frac{\left(2.3....49\right).\left(4.5....51\right)}{\left(3.4....50\right).\left(3.4.....50\right)}\)
\(A=\frac{2\left(3.4.....49\right).\left(4.5.....50\right).51}{\left(3.4.....49\right).50.3.\left(4.5.....50\right)}\)
\(A=\frac{2.51}{3.50}\)
\(A=\frac{2.17.3}{3.25.2}\)
\(A=\frac{17}{25}\)
EC
1 tháng 9 2016
Ta có:
P=12(52+1)(54+1)(58+1)(516+1)
P=(52-1)(52+1)(54+1)(58+1)(516+1):2
P=(54-1)(54+1)(58+1)(516+1):2
P=(58-1)(58+1)(516+1):2
P=(516-1)(516+1):2
P=(532-1):2
1 tháng 9 2016
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)=\frac{5^{32}-1}{2}\)
NH
8 tháng 11 2019
A = (62 + 72 + 82 + 92 + 102) - (12 + 22 + 32 + 42 + 52)
A = 62 + 72 + 82 + 92 + 102 - 12 - 22 - 32 - 42 - 52
A = (62 - 22) + (72 - 12) + (82 - 42) + (92 - 32) + (102 - 52)
A = (22 . 32 - 22) + (49 - 1) + (22 . 42 - 42) + (32 . 32 - 32) + (22 . 52 - 52)
A = 22(32 - 1) + 48 + 42(22 - 1) + 32(32 - 1) + 52(22 - 1)
A = (22 - 1)(42 + 52) + (32 - 1)(22 + 32) + 48
A = (4 - 1)(16 + 25) + (9 - 1)(4 + 9) + 48
A = 3 . 41 + 8 . 13 + 48
A = 123 + 104 + 48 = 275
HD
0
VT
13 tháng 4 2019
Điều kiện: X khác -1
\(A=\left(\frac{4}{x+1}-1\right):\frac{9+x^2}{x^2+2x+1}\)
\(=>A=\left(\frac{4}{x+1}-\frac{x+1}{x+1}\right):\frac{9+x^2}{\left(x+1\right)^2}\)
\(=>A=\frac{4-x-1}{x+1}\cdot\frac{\left(x+1\right)^2}{9+x^2}\)
\(=>A=\frac{\left(3-x\right)\cdot\left(x+1\right)}{9+x^2}\)
19 tháng 12 2020
a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)
\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)
b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{-6}{x-2}\)