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\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{14}{30}.\dfrac{15}{32}=\dfrac{1}{2^x}\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^x}\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{2\cdot4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^{x+1}}\)
\(\Rightarrow\dfrac{1}{2^{15}\cdot32}=\dfrac{1}{2^{x+1}}\)
\(\Rightarrow2^{15}.2^5=2^{x+1}\)
\(\Rightarrow2^{20}=2^{x+1}\)
\(\Rightarrow x+1=20\Rightarrow x=19\)
Vậy x = 19.
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
4) \(3^{n+2}+3^n=270\)
\(\Rightarrow3^n.3^2+3^n=270\)
\(\Rightarrow3^n.\left(3^2+1\right)=270\)
\(\Rightarrow3^n.\left(9+1\right)=270\)
\(\Rightarrow3^n.10=270\)
\(\Rightarrow3^n=270:10\)
\(\Rightarrow3^n=27\)
\(\Rightarrow3^n=3^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{15}{32}=2^x\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot32}=2^x\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{\left(2\cdot2\right)\left(2\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2.16\right)}=2^x\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot15}{2^{15}\cdot\left(2\cdot3\cdot4\cdot...\cdot16\right)}=2^x\Rightarrow\dfrac{1}{2^{19}}=2^x\)
\(\Rightarrow2^{-19}=2^x\Rightarrow x=-19\)
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{15}{32}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot15}{4\cdot6\cdot8\cdot...\cdot32}=\dfrac{1}{2^{14}\cdot2^5}=\dfrac{1}{2^{19}}=2^{-19}=2^x\\ \text{Vì }2\ne0;2\ne\pm1\text{ nên }x=-19\)