5x+5x+2=3125 | |
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\(a,TH1:x-2021=0=>x=2021\)
\(Th2:x-2022=0=>x=2022\)
Vậy \(x\in\left\{2021;2022\right\}\)
\(b,x\left(8-5\right)=1080\)
\(x.3=1080\)
\(x=360\)
\(c,x^3=216< =>6^3=216=>x=3\)
\(d,5^5=3125\)
a) ( x- 2021) * ( x- 2022) = 0
=> \(\orbr{\begin{cases}x-2021=0\\x-2022=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2021\\x=2022\end{cases}}}\)
b) b. 8x - 5x = 2022
=> 3x = 2022
=> x = 674
c) \(5\cdot x^3=1080\)
=> \(x^3=216\)
=> \(x^3=6^3\)
=> x = 6
d) \(5^x=3125\)
=> \(5^x=5^5\)
=> x = 5
\(\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(5x-1\right)^2-2\left(5x-1\right)\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2\)
\(=\left(5x-1-5x-4\right)^2\)
\(=\left(-5\right)^2\)
\(=25\)
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2
A = (5\(x\) + 1)2 + (5\(x\) - 1)2 - 2.( 5\(x\) +1).(5\(x\) - 1) tại \(x\) = 1
Thay \(x\) = 1 vào A ta có:
A = (5.1 + 1)2 + (5.1 - 1)2 - 2.(5.1 + 1).(5.1 - 1)
A = 62 + 42 - 2.6.4
A = 36 + 16 - 48
A = 52 - 48
A = 4
Lời giải:
Tại $x=4$ thì:
\(A=5(x^5-x^4+x^3-x^2+x-1)-1\)
\(=(x+1)(x^5-x^4+x^3-x^2+x-1)-1=x^6+1-1=x^6\)
\(=4^6=4096\)
5x+5x+2=3125
<=> 5x+5x.52= 3125
<=> 5x ( 1+52)= 3125
<=> 5x = 3125/26 <=> x = 2, 975630801
Ta có:
\(5^x+5^{x+2}=3125\)
\(\Leftrightarrow5^x+5^x\cdot5^2=3125\)
\(\Leftrightarrow5^x\left(1+25\right)=3125\)
\(\Leftrightarrow5^x=\frac{3125}{26}\)
\(\Leftrightarrow5^x\approx5^{2,975630801}\)
\(\Leftrightarrow x\approx2,975630801\)