4x+1+40= 65
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a) \(4x+12:3=40\)
\(4x+4=40\)
\(4x=40-4\)
\(4x=36\)
\(x=36:4\)
\(x=9\)
b) \(2^{x+7}+2017^0=65\)
\(2^{x+7}+1=65\)
\(2^{x+7}=65-1\)
\(2^{x+7}=64\)
\(2^{x+7}=2^6\)
\(x+7=6\)
\(x=6-7\)
\(x=-1\)
Lời giải chi tiết:
40 + 8 = 48 | 30 + 5 = 35 | 23 + 6 = 29 | 65 + 3 = 68 |
60 + 1 = 61 | 90 + 2 = 92 | 23 + 60 = 83 | 3 + 65 = 68 |
651<531+631+731+…+202331<40
165<153+163<153+163+173+…+120233<153+163+173+…+120233+…+120233651<531+631<531+631+731+…+202331<531+631+731+…+202331+…+202331
165<153+163<153+163+173<153+163+173+…+120233651<531+631<531+631+731<531+631+731+…+202331
Chúng ta có thể thấy rằng:
173+…+120233<165×(20233−73+1)731+…+202331<651×(20233−73+1)
173+…+120233<165×20161731+…+202331<651×20161
173+…+120233<165×311731+…+202331<651×311
173+…+120233<31165731+…+202331<65311
Từ đó, chúng ta có thể kết luận rằng:
165<153+163+173+…+120233<31165651<531+631+731+…+202331<65311
Vì 31165≈4.7846<4065311≈4.7846<40
=) đpcm
a) \(8x+56:14=60\)
\(\Rightarrow8x+4=60\)
\(\Rightarrow8x=56\)
\(\Rightarrow x=\dfrac{56}{8}\)
\(\Rightarrow x=7\)
b) Mình làm rồi nhé !
c) \(41-2^{x+1}=9\)
\(\Rightarrow2^{x+1}=41-9\)
\(\Rightarrow2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
d) \(3^{2x-4}-x^0=8\)
\(\Rightarrow3^{2x-4}-1=8\)
\(\Rightarrow3^{2x-4}=9\)
\(\Rightarrow3^{2x-4}=3^2\)
\(\Rightarrow2x-4=2\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
g) \(65-4^{x+2}=2014^0\)
\(\Rightarrow65-4^{x+2}=1\)
\(\Rightarrow4^{x+2}=64\)
\(\Rightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=1\)
i) \(120+2\left(4x-17\right)=214\)
\(\Rightarrow2\left(4x-17\right)=214-120\)
\(\Rightarrow2\left(4x-17\right)=94\)
\(\Rightarrow4x-17=47\)
\(\Rightarrow4x=47+17\)
\(\Rightarrow4x=64\)
\(\Rightarrow x=16\)
a: \(8x+56:14=60\)
=>8x+4=60
=>8x=60-4=56
=>x=56/8=7
b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)
=>2x-3=3
=>2x=6
=>x=3
c: \(41-2^{x+1}=9\)
=>\(2^{x+1}=41-9=32\)
=>x+1=5
=>x=4
d: \(3^{2x-4}-x^0=8\)
=>\(3^{2x-4}-1=8\)
=>\(3^{2x-4}=8+1=9\)
=>2x-4=2
=>2x=6
=>x=3
g: \(65-4^{x+2}=2014^0\)
=>\(65-4^{x+2}=1\)
=>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
i: 120+2(4x-17)=214
=>2(4x-17)=214-120=94
=>4x-17=94/2=47
=>4x=64
=>\(x=\dfrac{64}{4}=16\)
1: =>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)
=>2x-3=3
=>2x=6
=>x=3
2: \(41-2^{x+1}=9\)
=>\(2^{x+1}=32\)
=>x+1=5
=>x=4
3: =>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
\(5^{2x-3}-2.5^2=5^2.3\\ 5^{2x-3}=5^2.3+5^2.2\\ 5^{2x-3}=5^2.\left(3+2\right)\\ 5^{2x-3}=5^2.5\\ 5^{2x-3}=5^3\\ \Rightarrow2x-3=3\\ 2x=3+3\\ 2x=6\\ x=\dfrac{6}{2}\\ Vậy:x=3\)
4^(4x-5)-1=65
4^(4x-5)=65-1
4^(4x-5)=64
4^(4x-5)=4^3 (vì 4>0)
=>4x-5=3
=>4x=3+5
=>4x=8
=>x=8:4
=>x=2
Vậy x=2
\(4^{x+1}+4^0=65\)
\(4^{x+1}+1=65\)
\(4^{x+1}=65-1\)
\(4^{x+1}=64\)
\(4^{x+1}=4^3\)
=> X+1=3
x =2
Vậy x=2
\(4^{x+1}\) + \(4^0\) = 65
\(\Rightarrow\) \(4^{x+1}\) + 1 = 65
\(\Rightarrow\) \(4^{x+1}\) = 65 - 1 = 64 = \(4^3\)
\(\Rightarrow\) x + 1 = 3
\(\Rightarrow\) x = 3 - 1 = 2
Vậy x = 2 .