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a) \(45:\left(3x-19\right)=3^2\)
\(3x-19=45:9\)
\(3x=5+19\)
\(x=\frac{24}{3}\)
\(x=8\)
vậy \(x=8\)
b) \(57+7.64-x:3^2=500\)
\(448-x:9=500-57\)
\(448-x:9=443\)
\(x:9=448-443\)
\(x:9=5\)
\(x=45\)
vậy \(x=45\)
c) \(2^3.2^x=128\)
\(2^{x+3}=2^7\)
\(\Rightarrow x+3=7\)
\(\Rightarrow x=4\)
vậy \(x=4\)
a. 45:(3x-19)=3^2
45:(3x-19)=9
3x-19=45:9
3x-19=5
3x =5+19
3x =24
x =24:3
x =8
mấy câu còn lại mình chịu
Ta có: 2(x-5)-3(x-4)=-6+15(-3)
=>2x-10-3x+12=-6-45
=>-1x+2=-51
=>-1x=-53
=>x=53
Vậy x=53
Tìm x biết : 2 ( x - 5 ) - 3 ( x - 4 ) = - 6 + 15 ( - 3 )
2.(x-5)-3.(x-4)=-6+15.-3
2 (x − 5) − 3 (x − 4) = −51
(2x − 10) − (3x − 12) = −51
2x − 10 − 3x + 12 = −51
(2x − 3x) + (−10 + 12) = −51
−x + 2 = −51 −x = −53
x = 53
Vậy x = 53.
Bạn ơi, cái đó là 1,2x hay là 1x2x vậy bạn, bạn có thể giải thích cho mình được không, mk chưa rõ đề!!!
Ta có:
\(4\left(1+5+5^2+...+5^9\right)=5\left(1+5+5^2+...+5^9\right)-\left(1+5+5^2+...+5^9\right)\)
\(=5+5^2+5^3+...+5^{10}-1-5-5^2-...-5^9\)
\(=5^{10}-1+\left(5-5\right)+\left(5^2-5^5\right)+..+\left(5^9-5^9\right)\)
\(=5^{10}-1\)
=> \(1+5+5^2+...+5^9=\frac{5^{10}-1}{4}\)
Tương tự: \(1+5+5^2+...+5^8=\frac{5^9-1}{4}\)
\(1+3+3^2+...+3^9=\frac{3^{10}-1}{2}\)
\(1+3+3^2+...+3^8=\frac{3^9-1}{2}\)
=> \(A=\frac{5^{10}-1}{5^9-1}>\frac{5^{10}-1}{5^9}=5-\frac{1}{5^9}>4;\)
\(B=\frac{3^{10}-1}{3^9-1}< \frac{3^{10}}{3^9-1}=3+\frac{3}{3^9-1}< 4;\)
=> A > B.
2.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}.\left(\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow\)2x + 3 = 93
\(\Rightarrow\)2x = 93 - 3
\(\Rightarrow\)2x = 90
\(\Rightarrow\)x = 90 : 2 = 45
\(H=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{33.37}\)
= \(\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{37}\right)\)
= \(\frac{3}{4}\left(1-\frac{1}{37}\right)\)
= \(\frac{3}{4}.\frac{36}{37}=\frac{27}{37}\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(100x+\left(1+2+3+4+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700\div100\)
\(x=7\)
Vậy ...
( X + 1 ) + ( X + 2 ) + ( X + 3 ) + ... + ( X + 100 ) = 5750
100 X + ( 1 + 2 + 3 + ... + 100 ) = 5750
100 X + 5050 = 5750
100 X = 5750 - 5050
100 X = 700
X = 700 : 100
X = 7
Vậy x = 7
1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)
\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)
\(\Rightarrow3S-S=3^{2021}-1\)
\(\Rightarrow2S=3^{2021}-1\)
\(\Rightarrow S=\frac{3^{2021}-1}{2}\)
2. \(\left(3x-2\right)^3=64\)
\(\Leftrightarrow\left(3x-2\right)^3=4^3\)
\(\Leftrightarrow3x-2=4\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
[3x-2]^3=64
Ta có:64=4^3
Suy ra:3x-2=4
3x =4+2
3x=6
x=6:3
x=2