Tìm x
3 2 . 3 x = 81
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a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
2*(4x+1)+3*(2x+3)=81
8x+2+6x+9=81
14x+11=81
14x=81-11
14x=70
x=70:14
x=5
Dễ thấy (\(\frac{3}{4}\)-81); (\(\frac{3^2}{5}\)-81); (\(\frac{3^3}{6}\)-81);... (\(\frac{3^{2007}}{2010}\)-81) có dạng (\(\frac{3^x}{3+x}\)-81) và x\(\varepsilon\){1;2;3;...2007}.
Nếu x=6 thì \(\frac{3^x}{3+x}\)-81=\(\frac{3^6}{3+6}\)-81=0
=> (\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)=0
Mà |x-30|-6001=(\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)
=>|x-30|-6001=0
=>|x-30|=6001
=>x-30=6001 hoặc x-30=-6001
=>x=6031 hoặc x=-5971
-------------------The end----------------
\(\text{|x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2007}}{2010}-81\right)\)
\(\Rightarrow\text{ |x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)
\(\Rightarrow\left|x-30\right|- 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(3^4-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)
\(\Rightarrow|x - 30| - 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...0...\left(\frac{3^{2007}}{2010}-81\right)\)
\(\Rightarrow\text{|x - 30| - 6001 = }0\)
\(\Rightarrow\left|x-30\right|=6001\)
\(\Rightarrow x-30=6001\)hoặc \(x-30=-6001\)
\(\Rightarrow x=6031\)hoặc\(x=-5971\)
Vậy: x= 6031 hoặc x= -5971
(Nói thật thì mình mới lớp 7, đây có phải của lớp 8 không?)
1)
Ta có: \(\left(x-\frac{1}{2}\right)^3=8\Rightarrow\left(x-\frac{1}{2}\right)^3=2^3\)
\(\Rightarrow x-\frac{1}{2}=2\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}\)
\(\left(x-1\right)^3=\frac{8}{27}\Rightarrow\left(x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow x-1=\frac{2}{3}\Rightarrow x=\frac{2}{3}+1=\frac{5}{3}\)
a, => 3^x.(1+3+3^2)-1 = 1052
=> 3^x.13 = 1052+1 = 1053
=> 3^x = 1053 : 13
=> 3^x = 81 = 3^4
=> x = 4
b, => x^2-49 >=0 ; 81-x^2 >=0 hoặc x^2-49 < = 0 ; 81-x^2 < = 0
=> 49 < = x^2 < = 81
=> -9 < = x < = -7 hoặc 7 < = x < = 9
=> x thuộc {-9;-8;-7;7;8;9}
Tk mk nha
`3^{2}.3^{x}=81`
`=>3^{x+2}=3^{4}`
`=>x+2=4`
`=>x=2`
\(3^2\cdot3^x=81\)
\(3^{2+x}=81\)
\(3^{2+x}=3^4\)
\(\Rightarrow2+x=4\)
\(x=2\)