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\(E=2.3+3.4+4.5+3.6+2.7+4.15=2\left(3+7\right)+3\left(4+6\right)+4\left(5+15\right)=2.10+3.10+4.20=20+30+80=130\)
\(F=3\left(12+13+14+15\right)+3\left(8+7+6+5\right)=3\left(12+8+13+7+14+6+15+5\right)=3\left(20+20+20+20\right)=3.80=240\)
f: Ta có: \(E=3\cdot\left(12+13+14+15\right)+3\left(8+7+6+5\right)\)
\(=3\left(12+13+14+15+8+7+6+5\right)\)
\(=3\cdot80=240\)
720 : [ 41 - ( 2x - 5 ) ] = 23 . 5
720 : [ 41 - ( 2x - 5 ) ] = 40
[ 41 - ( 2x - 5 ) ] = 720 : 40
41 - ( 2x - 5 ) = 18
2x - 5 = 41 - 18
2x - 5 = 23
2x = 23 + 5
2x = 28
x = 28 : 2
x = 14
Vậy x = 14
a. 720 : [ 41 - ( 2x - 5 ) ] = 23 . 5
=> 720 : [ 41 - (2x-5) ] = 8.5
=> 720 : [ 41 - (2x-5) ] = 40
=> 41 - (2x-5) = 720 : 40
=> 41 - (2x-5) = 18
=> 2x - 5 = 41 - 18
=> 2x - 5 = 23
=> 2x = 23 + 5
=> 2x = 28
=> x = 28 : 2
=> x = 14 thuộc Z
vậy______
b. ( 3x - 24 ) . 74 = 2.75
=> 3x-16 = 2.75 : 74
=> 3x - 16 = 2.7
=> 3x - 16 = 14
=> 3x = 14 + 16
=> 3x = 30
=> x = 30 : 3
=> x = 10 thuộc Z
vậy_____
Ta có:
\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)
\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)
\(7^{2x+1}-2.7^4=7^4.5\)
\(\Rightarrow7^{2x+1}=7^4.5+2.7^4\)
\(\Rightarrow7^{2x+1}=7^4.\left(5+2\right)\)
\(\Rightarrow7^{2x+1}=7^4.7\)
\(\Rightarrow7^{2x+1}=7^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
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