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a) 2x = 16 <=>x=8
b) 3x+1 = 9x <=>9x-3x=1
<=>6x=1 <=>x=1/6
c) 23x+2 = 4x+5 <=>23x-4x=5-2
<=>19x=3 <=>x=3/19
d) 32x-1 = 243 <=>32x=244
<=>x=61/8
a/ 2x=16
x=8
b/ 3x+1=9x
3x-9x=-1
-6x=-1
x=1/6
c/ 23x+2=4x
23x-4x=-2
19x=-2
x=-2/19
d/ 32x-1=243
32x=244
x=61/8
\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)
Bài 1:
a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)
b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)
c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)
d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)
Bài 2:
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
Bài 3:
a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)
b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)
`C = (3x-15-5)/(x-5)`
`= 3 - 5/(x-5) in ZZ <=> 5/(x-5) in ZZ => 5 vdots x-5`.
`=> x - 5 in Ư(5)`.
`<=> x in {0; 10; 4; 6}`
Vậy...
23x + 2 = 4x + 5
=> 23x - 4x = 5 - 2
=> 19x = 3
=> x = 3/19
23x+2=4x+5
23x-4x=5-2
19x=3
x=3:19
x=\(\frac{3}{9}\)
Vậy x=\(\frac{3}{9}\)là số cần tìm
\(F=\frac{3}{2}x^4-\frac{1}{16}x^4+\frac{1}{32}x^4-\frac{1}{4}x^4\)
\(F=\left(\frac{3}{2}-\frac{1}{16}+\frac{1}{32}-\frac{1}{4}\right)x^4\)
\(F=\frac{39}{32}x^4\)
Ta có : x4 có số mũ là 4 => x4 luôn dương với mọi x ( x khác 0 )
\(\frac{39}{32}>1\Rightarrow\frac{39}{32}>0\)
=> \(\frac{39}{32}x^4\)luôn dương với mọi x ( x khác 0 )
=> \(\frac{39}{32}x^4>0\)với mọi x ( x khác 0 )
=> \(F=\frac{3}{2}x^4-\frac{1}{16}x^4+\frac{1}{32}x^4-\frac{1}{4}x^4>0\forall x\left(x\ne0\right)\)
cho c xl nha em :( c ko bt em la ai thật
Em không bạn tâm đâu!