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TL :
120 : 54 - [ 50 : 2 - 32 - 2 x 4 ]
120 : 54 - [ 50 : 2 - 9 - 8 ]
120 : 54 - [ 25 - 1 ]
120 : 54 - 24
120 : 30
4
120/{54-[50/2-(3^2-2*4)]}
120/{54-[50/2-(9-8)]}
120/{54-[25-1]}
120/{54-24}
120/30
4
\(=\left(-3\right)^2\left(135+130\right)+5=9.265+5=2385+5=2390\)
N = 120 . 20 = 2400 nu
A + T = 60% => A=T = 30%
G= X = 50% - 30% = 20%
A=T = 2400 x 30% = 720 nu
G=X= 2400 x 20% = 480 nu
a: \(=\dfrac{6\left(12-7\right)}{60}=\dfrac{6\cdot5}{6\cdot10}=\dfrac{5}{10}=\dfrac{1}{2}\)
b: \(=\dfrac{35\cdot\left(18-1\right)}{34\cdot7}=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\)
c: \(=\dfrac{42\left(1-11\right)}{21\cdot\left(-15\right)}=2\cdot\dfrac{-10}{-15}=2\cdot\dfrac{2}{3}=\dfrac{4}{3}\)
d: \(=\dfrac{2^5\cdot3^2}{2^5\cdot3^3}=\dfrac{1}{3}\)
e: \(=\dfrac{-36\cdot14}{27\left(14+7\right)}=\dfrac{-36}{27}\cdot\dfrac{14}{21}=\dfrac{-4}{3}\cdot\dfrac{2}{3}=-\dfrac{8}{9}\)
a) (x- 35 ) - 100 = 0
(x- 35 ) = 0 + 100
(x- 35 ) = 100
x = 100 + 35
x = 135
b) 120 + ( 118 - x ) = 213
( 118 - x ) = 213-120
( 118 - x ) = 93
x = 118 - 93
x = 25
c) 150 - ( x + 60 ) = 76
( x + 60 ) = 150 - 76
( x + 60 ) = 74
x = 74 - 60
x = 14
a) ( x - 35 ) - 100 = 0
-> x - 35 - 100 = 0
-> x - 135 = 0
-> x = 135
b) 120 + ( 118 - x ) = 213
-> 118 - x = 213 - 120
-> 118 - x = 93
-> x = 118 - 93
-> x = 25
c) 150 - ( x + 60 ) = 76
-> x + 60 = 150 - 76
-> x + 60 = 74
-> x = 14
CHÚC BẠN HỌC TỐT !!! >...<
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)
\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)
\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)
\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)
\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)
A=120:{60:[(3mux2 + 4mux2)-5]}
A=120:{60[(9+16)-5]}
A=120:{60[25-5]}
A=120:{60.20}
A=120:1200
A=0,1
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