Tìm x , biết
3x +3x-2 = 810 với x ≥ 2
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\(3x\left(x+4\right)-3x^2-4=0\\ \Rightarrow3x^2+12x-3x^2-4=0\\ \Rightarrow12x-4=0\\ \Rightarrow12x=4\\ \Rightarrow x=\dfrac{1}{3}\)
\(3^x.3^2.3=243.3\\ \Rightarrow3^x.3^2=243\\ \Rightarrow3^x.3^2=3^5\\ \Rightarrow3^x=3^5:3^2\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
\(3x^2y^3-x^2y-M=x^2y^3+x^2y\\ \Rightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y\\ \Rightarrow M=2x^2y^3-2x^2y\)
\(\Leftrightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y=2x^2y^3-2x^2y\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
thay x=2k;y=3k;z=5k vào x.y.z=810 ta được:
2k.3k.5k=810
30.k3=810
k3=27
=>k=3
=>x=2.3=6
y=3.3=9
z=5.3=15
a) 3x = 2y \(\Rightarrow\)\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{2}.\frac{1}{5}=\frac{y}{3}.\frac{1}{5}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{3}=\frac{z}{7}.\frac{1}{3}\Rightarrow\frac{y}{15}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x+y+z}{10+15+21}=\frac{32}{46}=\frac{2}{3}\)
\(\hept{\begin{cases}x=10.\frac{2}{3}=\frac{20}{3}\\y=15.\frac{2}{3}=10\\z=21.\frac{2}{3}=14\end{cases}}\)
Vậy \(\hept{\begin{cases}x=10.\frac{2}{3}=\frac{20}{3}\\y=15.\frac{2}{3}=10\\z=21.\frac{2}{3}=14\end{cases}}\)
\(3^x+3^{x-2}=810\)
<=> \(3^x+3^x:3^2=810\)
<=> \(3^x\left(1+\frac{1}{9}\right)=810\)
<=> \(3^x=729\)
<=>\(3^x=3^6\)
=> x=6
3x +3x-2 = 810
\(\Rightarrow3^x\left(1+3^{-2}\right)=810\)
\(\Rightarrow3^x\left(1+\frac{1}{3^2}\right)=810\)
\(\Rightarrow3^x\left(1+\frac{1}{9}\right)=810\)
\(\Rightarrow3^x\cdot\frac{10}{9}=810\)
\(\Rightarrow3^x=729\)
\(\Rightarrow3^x=3^6\Leftrightarrow x=6\)