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5^217>119^72
2^1050>5^450
1000^10+1990^9>1991^10
(nếu đúng tki tích cho mk nha)
Bài 1 so sánh
333444 và 444333
Bài 2 so sánh
a) 321 và 231
b) 1990 10 + 19909 và 199110
c) 10750 và 7375
333444 và 444333
ta có : 333444 = ( 3334 )111 =12296370321111
444333 = ( 4443 )111 = 87528384111
vì 12296370321 > 87528384
=> 333444 > 444333
Bài 1:
333^444 = (333^4)^111
444^333 = (444^3)^111
333^4 = 3^4.111^4 = 81.111^4
444^3 = 4^3.111^3 = 64.111^3 < 81.111^4
444^3< 333^4
(333^4)^111 > (444^3)^111
333^444 > 444^333
Lời giải:
$1990^{10}+1990^9=1990^9(1990+1)=1991.1990^9< 1991.1991^9=1991^{10}$
-----------------------
$10^{10}=(10^2)^5=100^5=(2.50)^5=2^5.50^5=32.50^5< 48.50^5$
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$11^{1979}< 11^{1980}=(11^3)^{660}=1331^{660}$
$37^{1320}=(37^2)^{660}=1369^{660}> 1331^{660}$
$\Rightarrow 11^{1979}< 37^{1320}$
Bài 1:
a: \(10^{10}=\left(2\cdot5\right)^{10}=2^{10}\cdot5^{10}=2^9\cdot5^{10}\cdot2\)
\(48\cdot50^5=2^4\cdot3\cdot\left(2\cdot5^2\right)^5=2^4\cdot3\cdot2^5\cdot5^{10}=2^9\cdot5^{10}\cdot3\)
mà 2<3
nên \(10^{10}<48\cdot50^5\)
b: \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9\cdot1991\)
\(1991^{10}=1991^9\cdot1991\)
mà 1990<1991
nên \(1990^{10}+1990^9<1991^{10}\)
c: \(107^{50}<108^{50}=\left(2^2\cdot3^3\right)^{50}=2^{100}\cdot3^{150}\)
\(73^{75}>72^{75}=\left(2^3\cdot3^2\right)^{75}=2^{225}\cdot3^{150}\)
mà \(2^{225}\cdot3^{150}>2^{100}\cdot3^{150}=108^{50}>107^{50}\)
nên \(73^{75}>107^{50}\)
d: \(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
mà 8192>3125
nên \(2^{91}>5^{35}\)
e: \(A=72^{45}-72^{44}=72^{44}\left(72-1\right)=72^{44}\cdot71\)
\(B=72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)
mà 44>43
nên A>B
Bài 2:
a:
ĐKXĐ: x<>2023
\(\frac{x-2023}{4}=\frac{1}{x-2023}\)
=>\(\left(x-2023\right)\left(x-2023\right)=4\cdot1\)
=>\(\left(x-2023\right)^2=4\)
=>\(\left[\begin{array}{l}x-2023=2\\ x-2023=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+2023=2025\left(nhận\right)\\ x=-2+2023=2021\left(nhận\right)\end{array}\right.\)
b: \(\left(2x+1\right)^4=\left(2x+1\right)^6\)
=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)
=>\(\left(2x+1\right)^4\cdot\left\lbrack\left(2x+1\right)^2-1\right\rbrack=0\)
=>\(\left(2x+1\right)^4\cdot\left(2x+1-1\right)\left(2x+1+1\right)=0\)
=>\(2x\left(2x+1\right)^4\cdot\left(2x+2\right)=0\)
=>\(\left[\begin{array}{l}2x=0\\ 2x+1=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\\ x=-1\end{array}\right.\)
c: \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
=>\(\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
=>\(\left(3x-1\right)^{10}\cdot\left\lbrack\left(3x-1\right)^{10}-1\right\rbrack=0\)
=>\(\left[\begin{array}{l}\left(3x-1\right)^{10}=0\\ \left(3x-1\right)^{10}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-1=0\\ \left(3x-1\right)^{10}=1\end{array}\right.\)
=>\(\left[\begin{array}{l}3x-1=0\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=\frac23\\ x=0\end{array}\right.\)
d: Sửa đề \(2^{x+1}\cdot3^{y}=12^{x}\)
=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)
=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)
bài 1:
Ta có: 199010+19909=19909. 1990+19909=19909.1991
199110=19919. 1991
mà 19909<19919 nên 199010+19909<199110.