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Bài 1:
a) 02002 < 02023
b) 20220 = 20230
c) 549 < 5510
d) ( 4 + 5 )3 > 42 + 52
đ) 92 - 32 > ( 9 - 3 )2
Bài 2:
a) 32 x 43 - 32 + 333
= 9 x 64 - 9 + 333
= 576 - 9 + 333
= 567 + 333
= 900
b) 5 x 43 + 24 x 5 + 410
= 5 x 64 + 24 x 5 + 1
= 5 x ( 64 + 24 ) + 1
= 5 x 88 + 1
= 440 + 1
= 441
c) 23 x 42 + 32 x 5 - 40 x 12023
= 8 x 16 + 9 x 5 - 40 x 1
= 128 + 45 - 40
= 133
Bài 1 :
a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)
b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)
c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)
d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)
đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)
So sánh : và \(72^{44}-72^{43}\)
Ta có :
\(72^{45}-72^{44}=72^{44}\left(72-1\right)\)
\(72^{44}-72^{43}=72^{43}\left(72-1\right)\)
Vì 7244 > 7243 => 7244 (72-1) > 7243 (72-1)
hay 7245 -7244 > 7244 - 7243
a/ \(27^{11}=\left(3^3\right)^{11}=3^{33}\); \(81^8=\left(3^4\right)^8=3^{32}< 3^{33}\Rightarrow81^8< 27^{11}\)
b/ \(3^{2n}=\left(3^2\right)^n=9^n\); \(2^{3n}=\left(2^3\right)^n=8^n< 9^n\Rightarrow2^{3n}< 3^{2n}\)
a. 2711= (33)11 = 333
818 = (34)8 = 332
Suy ra 333>332 hay 2711>818
b. 32n = (32)n = 9n
23n = (23)n = 8n
Mà 9>8 suy ra 9n>8n hay 32n>23n
c. 523 = 522 . 5
(6.5)22 = 622 . 522
Vì 622>5 suy ra 522 . 5<622 . 522 hay 523<(6.5)22
d. 7245-7244 = 7244(72-1) = 7244 . 71
7244-7243 = 7243(72-1) = 7243 . 71
Vì 7244>7243 suy ra 7244 . 71>7243 . 71 hay 7245-7244>7244-7243
Sửa đề: Tìm x
a: \(x\left(6-x\right)^{2023}=\left(6-x\right)^{2023}\)
=>\(x\left(6-x\right)^{2023}-\left(6-x\right)^{2023}=0\)
=>\(\left(6-x\right)^{2023}\left(x-1\right)=0\)
=>\(\left[\begin{array}{l}6-x=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=1\end{array}\right.\)
b: \(5^{x}+5^{x+2}=650\)
=>\(5^{x}+5^{x}\cdot25=650\)
=>\(5^{x}\left(1+25\right)=650\)
=>\(5^{x}=\frac{650}{26}=25=5^2\)
=>x=2
c: \(2^{x+2}-2^{x}=96\)
=>\(2^{x}\cdot2^2-2^{x}=96\)
=>\(2^{x}\left(2^2-1\right)=96\)
=>\(2^{x}\cdot3=96\)
=>\(2^{x}=\frac{96}{3}=32=2^5\)
=>x=5
d: \(10^{x}:5^{y}=20^{y}\)
=>\(10^{x}=20^{y}\cdot5^{y}=100^{y}=\left(10\right)^{2y}\)
=>x=2y
b.Ta có : = (111.3)111.4 = ( 1114 . 34 )111=
= ( 111 . 4 )111.3 = ( 1113.43)111 =
Vì (1114.81)111 > ( 1113.64 )111 => 333444 > 444333
a. 1030 = ( 103 )10=100010
2100 = ( 210 )10=102410
Vì 100010<102410 nên 1030<2100
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
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 \(2^{x+1}\cdot3^{y}=12^{x}\)
=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)
=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)