Tính:
(2/5)7 . 57 + (9/4)3 : (3/16)3 / 27 . 52 + 512
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\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)
BT1 :thực hành phép tính
4/3+-11/31+3/10-20/31-2/5 = 7/30
7/12-1/-16+3/4 = 67/48
2/11.-5/4+-9/11.5/4+ 1 và 3/4 = -1/4 * và 3/4 là sao bạn *
-9/27-25/75 = -2/3
a) 1/2 + 3/4 - (3/4 - 4 - 5)
= 1/2 + 3/4 - 3/4 + 4 + 5
= (3/4 - 3/4) + (4 + 5) + 1/2
= 0 + 9 + 1/2
= 19/2
b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)
= 9/16 - 8/27 - 19/27 + 7/16 + 2
= (9/16 + 7/16) + (-8/27 - 19/27) + 2
= 1 - 1 + 2
= 2
c) -5/8 . [4/9 + 7/(-12)]
= -5/8 . (-5/36)
= 25/288
d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)
= 7/10 . (-3/5 - 2/5) + 3/10
= 7/10 . (-1) + 3/10
= -2/5
e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7
= -3/7 . (5/9 + 4/9) + 17/7
= -3/7 . 1 + 17/7
= 2
f) 8 2/7 - (3 4/9 + 4 2/7)
= 8 + 2/7 - 3 - 4/9 - 4 - 2/7
= (8 - 3 - 4) + (2/7 - 2/7) - 4/9
= 1 - 4/9
= 5/9
h) 3.(-1/2)² - (4/5 + 8/15) : 5/6
= 3.1/4 - 4/3 : 5/6
= 3/4 - 8/5
= -17/20
\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)=\(\frac{3^6.3^8.5^4-5^{13}.3^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3^{14}.5^4-5^4.3^{13}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3.1.}{1.5^2.}\)=\(\frac{3}{25}\)
Học tốt
50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)