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\(\frac{1}{27}.\left(3^x\right)^2=\frac{1}{243}\Rightarrow\left(3^x\right)^2=\frac{1}{243}:\frac{1}{27}\Rightarrow\left(3^x\right)^2=\frac{1}{9}\)
\(\Rightarrow3^x=\sqrt{\frac{1}{9}}\Rightarrow3^x=\frac{1}{3}\Rightarrow3^x=3^{-1}\Rightarrow x=-1\)
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Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
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Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}=\frac{50-5}{9}=5\)
\(\Rightarrow\hept{\begin{cases}x=5.2+1=11\\y=5.3+2=17\\z=5.4+3=23\end{cases}}\)
Vậy...
Ta có: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}2x−1=3y−2=4z−3
\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}⇒42x−2=93y−6=4z−3
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}=\frac{50-5}{9}=542x−2=93y−6=4z−3=4+9−42x+3y−z−2−6+3=950−5=5
\(\Rightarrow\hept{\begin{cases}x=5.2+1=11\\y=5.3+2=17\\z=5.4+3=23\end{cases}}\)
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\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
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Ta có:
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right).....\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right).\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right).....\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.-2.-3......-2002}{2.3.4.....2003}=\frac{1}{2003}\)
\(\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot......\cdot\left(\frac{1}{2003}-1\right)\)
=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.........\cdot\frac{2002}{2003}\) = \(\frac{1}{2003}\)
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3^2.1/243.81^2.1/3^2
=9/243.81^2/3^2
=1/27.27^2/1
=27^2/27
=27
\(3^2\times\frac{1}{243}\times81^2\times\frac{1}{3^3}\)
Ta có :
\(=9\times\frac{1}{243}\times6561\times\frac{1}{27}\)
\(=\frac{1}{27}\times6561\times\frac{1}{27}\)
\(=243\times\frac{1}{27}\)
\(=9\)
ủng hộ mik nhé .
\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
\(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}\)
\(=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}\)
\(=3^2.3^8.\frac{1}{3^5}.\frac{1}{3^3}\)
\(=3^{10}.\frac{1}{3^8}\)
tíc mình nha