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Bài 1 :
a. \(\left|x-\frac{1}{3}\right|< \frac{5}{2}\)
TH1 : nếu \(\left|x-\frac{1}{3}\right|>0\)
\(x-\frac{1}{3}< \frac{5}{3}\)
\(x< 2\)
TH2 : nếu \(\left|x-\frac{1}{3}\right|< 0\)
\(\frac{1}{3}-x< \frac{5}{3}\)
\(x>-\frac{4}{3}\)
Bài 2 :
a. \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2-1=0\)
\(\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
\(3^2\times\frac{1}{243}\times81^2\times\frac{1}{3^3}\)
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 .
\(\frac{25}{5^x}=\frac{1}{125}\Rightarrow25.125=5^x.1\)
\(3125=5^x\)
\(5^5=5^x\)
\(\Rightarrow x=5\)
a) x : \(\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(x:\frac{-1}{27}=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\frac{-1}{27}\)
\(x=\frac{1}{81}\)
Vậy \(x=\frac{1}{81}\)
a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(\Leftrightarrow x=\left(-\frac{1}{3}\right)\cdot\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow x=\left(-\frac{1}{3}\right)^4\)
\(\Leftrightarrow x=\frac{1}{81}\)
b)\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)
\(\Leftrightarrow x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=-\frac{1}{4}\)
d)\(\left(3x+1\right)^3=-27\)
\(\Leftrightarrow3x+1=-3\)
\(\Leftrightarrow3x=-4\)
\(\Leftrightarrow x=-\frac{4}{3}\)
\(x:y:z=3:4:5\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\) và \(5z^2-3x^2-2y^2\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{5z^2-3x^2-2y^2}{5.5^2-3.3^2-2.4^2}=\frac{594}{66}=9\)
\(\Leftrightarrow\frac{x}{3}=9\Rightarrow x=9.3=27\)
\(\Leftrightarrow\frac{y}{4}=9\Rightarrow y=9.4=36\)
\(\Leftrightarrow\frac{z}{5}=9\Rightarrow z=9.5=45\)
Vậy x = 27 ; y = 36 ; z = 45
\(x+y=3\left(x-y\right)\)
\(\Rightarrow x+y=3x-3y\)
\(\Rightarrow y+3y=3x-x\)
\(\Rightarrow4y=2x\)
\(\Rightarrow2y=x\)
\(\Rightarrow x:y=2\)
\(\Rightarrow x+y=2y+y=2\)
\(\Rightarrow3y=2\)
\(\Rightarrow y=\frac{2}{3}\)
\(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3};y=\frac{2}{3}\)
a) Ta có: \(\left[\frac{3}{7}\cdot\frac{4}{15}+\frac{1}{3}\cdot\left(9^{15}\right)\right]^0\cdot\frac{1}{3}\cdot\frac{6^{12}}{12^4}\)
\(=\frac{1}{3}\cdot\frac{6^{12}}{6^4\cdot2^4}=\frac{6^{12}}{6^4\cdot48}=\frac{\left(6^4\right)^3}{6^4\cdot48}=\frac{6^8}{48}=34992\)
b) Ta có: \(\frac{10^2\cdot81-16\cdot15^2}{4^4\cdot675}=\frac{2^2\cdot5^2\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)
\(=\frac{2^25^23^2\left(3^2-2^2\right)}{\left(2^2\right)^4\cdot3^3\cdot5^2}=\frac{\left(3^2-2^2\right)}{64\cdot3}=\frac{5}{192}\)
\(=\left(-\frac{1}{3}\right)^3=\frac{-1}{27}\)
\(\left(-\frac{1}{3}\right)^2.\left(-\frac{1}{3}\right)\)
\(=\frac{1}{9}.\left(-\frac{1}{3}\right)\)
\(=-\frac{1}{27}\)