Tìm x biết
2017 mũ 5 : ( x-60)=2017 mũ 6
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\(\left(x+1\right)^3=27\)
\(\left(x+1\right)^3=3^3\)
\(\Rightarrow x+1=3\)
\(x=2\)
\(\left(x+1\right)^3=27\)
\(< =>\left(x+1\right)^3=3.3.3=3^3\)
\(< =>x+1=3< =>x=3-1=2\)
\(\left(2x+3\right)^3=9.81\)
\(< =>\left(2x+3\right)^3=9.9.9\)
\(< =>\left(2x+3\right)^3=9^3\)
\(< =>2x+3=9< =>2x=6\)
\(< =>x=\frac{6}{2}=3\)
a: Ta có: \(60-3\left(x-2\right)=51\)
\(\Leftrightarrow3\left(x-2\right)=9\)
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=2^5:2^3\)
\(\Leftrightarrow4x=24\)
hay x=6
1. a) 4.415.8.25.125
= (4.25). (8.125).415
= 100.1000.415
= 100000.415
= 41500000
b) 2.31.12+4.42.6+8.27.3
= (2.31.12)+(4.42.6)+(8.27.3)
= (2.12).31+(4.6).42+(8.3).27
= 24.31+24.42+24.27
= 24 (31+42+27)
= 24.100
= 2400
(59 x 75 - 510 x 75 : 5) : 20172018
= [59 x 75 - (510 : 5) x 75) : 20172018
= (59 x 75 - 59 x 75) : 20172018
= 0 : 20172018
= 0
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
a) x=3
b) x=1
c) x=1 hoặc -5
d) x=2
e) x=2
g) x=2
h) x=1 hoặc x=0 hoặc x=-1
i) x=-1 hoặc x=0
\(a.4^x=64\)
\(4^x=4^3\)
\(\Rightarrow x=3\)
\(b,3^{x\times4}=81\)
\(3^{x\times4}=3^4\)
\(x\times4=4\)
\(\Rightarrow x=1\)
\(c,\left(2+x\right)^4=81\)
\(\left(2+x\right)^4=3^4\)
\(2+x=3\)
\(x=3-2\)
\(x=1\)
\(d,5^{x\times5}=125\)
\(5^{x\times5}=5^3\)
\(x\times5=3\)
\(x=3:5\)
\(x=\frac{3}{5}\)
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
20175:(x-60)=20176
\(\Leftrightarrow\)x-60 =20176-5
\(\Leftrightarrow\)x-60 =2017
\(\Leftrightarrow\)x =2017+60
\(\Leftrightarrow\)x =2077 (TMĐK)
Vậy x =2077
20175 :(x-60)=20176
x-60=20175 :20176 =1/2017
x=1/2017+60=121021/2017