b) 56 + ( x + 1 )3 = 120 ( Tìm x )
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[(6x - 72) : 2 - 84]. 28 = 5628
=> (6x - 72) : 2 - 84 = 5628 : 28 = 201
=> (6x - 72) : 2 = 201 + 84 = 285
=> 6x - 72 = 285 . 2 = 570
=> 6x = 570 + 72 = 642
=> x = 642 : 6
=> x = 107
\(x\in BC\left(20;120;56\right)\)
\(\Leftrightarrow x=840\)
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
=>\(x=\dfrac{64}{4}=16\)
`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
a, 15 : x + 120 : 12 =15
\(\Rightarrow\) 15 : x + 10 = 15
\(\Rightarrow\) 15 : x = 15 - 10
\(\Rightarrow\) 15 : x = 5
\(\Rightarrow\) x = 15 : 5
\(\Rightarrow\) x = 3
b, [ ( x + 34 ) - 50].2 = 56
\([\left(x+34\right)-50]=56:2\)
(x + 34 ) - 50 = 28
x + 34 = 28+50
x+34 =78
x = 78 - 34
x = 44
a,15:x+120 : 12 = 15
15 : x + 10 = 15
15: x = 15 -10
15 : x = 5
x = 15 : 5
x = 3
vậy x = 3
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
`56 + (x+1)^3 =120`
`(x+1)^2 = 120 -56 = 64`
`=> x+1=4`
`x =4-1=3`
Vậy `x=3`
56+(x+1)3=120
=>(x+1)3=120-56
=>(x+1)3=64<=>64=4.4.4=43
=>(x+1)3=43
=>x+1=4
=>x=4-1
=>x=3
Vậy x=3