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1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
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\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)
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a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
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1/ \(2x^3=2^4\)
\(\Rightarrow x^3=2^4:2\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
tíc mình nha
1/ 2 . x3 = 24 => x3 = 24 : 2 => x3 = 23 => x = 2
2/ Ta có: Số số hạng của X là: (87 - 3) : 3 + 1 = 29 (số hạng)
=> X = 3 + 6 + 9 + ..... + 87 = (87 + 3) x 29 : 2 = 1305
3/ A = { 50 ; 52 ; 54 ; 56 ; 58 ; 60 }
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1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x
(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x
(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x
(31 số 1/2)
(1/2)^31. = 2^x
=> 0=x+36
x=0-36
x=-36
Vậy x=-36
Theo mk nghĩ,mk làm đúng nha .Tk cho mk
Để mk sửa phần này một chút
\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)
\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)
\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)
\(\frac{1}{2^{36}}=2^x\)
\(1=2^x\cdot2^{36}\)
\(2^0=2^x+36\)
Rồi bn tự suy luận nha
Trên ảnh là rose mà , jisoo khác
\(2^{x+4}-2^{x+3}-2^{x+2}-2^{x+1}-2=62\)
\(\Rightarrow\) \(2^x.2^4-2^x.2^3-2^x.2^2-2^x.2^1=64\)
\(\Rightarrow\) \(2^x.\left(2^4-2^3-2^2-2^1\right)=64\)
\(\Rightarrow\) \(2^x.\left(16-8-4-2\right)=64\)
\(\Rightarrow\) \(2^x.2=64\)
\(\Rightarrow\) \(2^x=32\)
\(\Rightarrow\) \(2^x=2^5\)
\(\Rightarrow\) \(x=5\)