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x10 = 25x8
⇒ x10 − 25x8 = 0
⇒ x8.(x2 − 25) = 0
Suy ra x8 = 0 hoặc x2 - 25 = 0.
Do đó x = 0 hoặc x = 5 hoặc x = -5.
Vậy x ∈ {0; 5; −5}.
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Lời giải:
$M=(x^{10}-24x^9)-(x^9-24x^8)+(x^8-24x^7)-(x^7-24x^6)+(x^6-24x^5)-(x^5-24x^4)+(x^4-24x^3)-(x^3-24x^2)+(x^2-24x)-(x-24)+1$
$=x^9(x-24)-x^8(x-24)+x^7(x-24)-.....+x(x-24)-(x-24)+1$
$=(x-24)(x^9-x^8+x^7-...+x-1)+1$
$=0.(x^9-x^8+....+x-1)+1=1$
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\(M=x^{10}-25x^9+25x^8-25x^7+...-25x^3+25x^2-25x+25\)
Ta thấy : \(x=24\Rightarrow x+1=25\)
\(\Rightarrow M=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)
\(M=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(\Rightarrow M=1\)
Vậy \(M=1\left(tạix=24\right)\)
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Ta có: \(x^8\cdot25^4=x^{10}\)
\(\Leftrightarrow x^{10}:x^8=625^2\)
\(\Leftrightarrow x^2=625^2\)
hay \(x\in\left\{625;-625\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có: x=100
\(\Leftrightarrow x+1=101\)
Ta có: \(f\left(x\right)=x^{10}-101x^9+101x^8-101x^7+...+101x+2021\)
\(=x^{10}-x^9\cdot\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x\left(x+1\right)+2021\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^2+x+2021\)
\(=x+2021\)
\(=100+2021=2121\)
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\(\left(0,1\right)^4.\left(0,1\right)^2.10^4\)
\(=\frac{1}{10^4}.10^4.\frac{1}{10^2}\)
\(=1.\frac{1}{100}=\frac{1}{100}\)
Ta có :
(0,1)4 x (0,1)2 x 104
=( 0,1 x 10)4 x 0,01
= 14 x 0,01
= 0,01
\(x^{10}-25x^8=0\) => \(x^8.\left(x^2-25\right)=0\)
=> \(x^8=0\) => x=0
hoặc \(x^2-25=0\)=> x= -5; 5
Vậy \(x=\left\{-5;0;5\right\}\)
Ta có:x\(^{10}\)-25.x\(^8\)=0
x\(^8\).x\(^{^2}\)- 25. x\(^8\)=0
x\(^8\)(x2-25)=0