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a, \(\left(2x-1\right)^2:9=49\)
\(\left(2x-1\right)^2=441\)
\(\Rightarrow\orbr{\begin{cases}2x-1=441\\2x-1=-441\end{cases}\Rightarrow\orbr{\begin{cases}x=221\\x=-220\end{cases}}}\)
b, \(3^x+3^{x+2}=810\)
\(3^x+3^x.3^2=810\)
\(3^x\left(1+3^2\right)=810\)
\(3^x.10=810\)
\(3^x=81=3^4\)
\(\Rightarrow x=4\)
\(\hept{\begin{cases}80⋮x\\56⋮x\end{cases}}\Rightarrow x\inƯC\left(80;56\right)\)
\(80=2^4.5\)
\(56=2^3.7\)
\(ƯCLN\left(80;56\right)=2^3=8\)
\(\RightarrowƯC\left(80;56\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Mà \(x\ge3\)
\(\Rightarrow x\in\left\{4;8\right\}\)
a)3/7x=8/13y=6/19z và 2x-y-z =-6
b)x/8=y/3=7/10 va xy+yz+zx=1206
c) x/4=2y/5=5z/6 và x2- 3y2+2z2=325
a) \(\frac{3}{7}x=\frac{8}{13}y=\frac{6}{19}z\) và 2x-y-z =-6
=)\(\frac{x}{\frac{7}{3}}=\frac{y}{\frac{13}{8}}=\frac{z}{\frac{19}{6}}=\frac{2x}{\frac{14}{3}}=\frac{y}{\frac{13}{8}}=\frac{z}{\frac{19}{6}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2x}{\frac{14}{3}}=\frac{y}{\frac{13}{8}}=\frac{z}{\frac{19}{6}}=\frac{2x-y-z}{\frac{14}{3}-\frac{13}{8}-\frac{19}{6}}=\frac{-6}{\frac{-3}{24}}=48\)
\(\Rightarrow\frac{x}{\frac{7}{3}}=48\Rightarrow x=48\times\frac{7}{3}=112\)
\(\Rightarrow\frac{y}{\frac{13}{8}}=48\Rightarrow y=48\times\frac{13}{8}=78\)
\(\Rightarrow\frac{z}{\frac{19}{6}}=48\Rightarrow z=48\times\frac{19}{6}=152\)
Vậy x=112;y=78;z=152
\(9^x:3^{x+9}=27\)
\(9^x=27\cdot3^{x+9}\)
\(\left(3^2\right)^x=3^3\cdot3^{x+9}\)
\(3^{2x}=3^{x+12}\)
\(\Rightarrow2x=x+12\)
\(2x-x=12\)
\(x=12\)
\(4^{x+y}:2^{5y}=32\)
\(4^{12+y}=32\cdot2^{5y}\)
\(\left(2^2\right)^{12+y}=2^5\cdot2^{5y}\)
\(2^{24+2y}=2^{5+5y}\)
\(24+2y=5+5y\)
\(24-5=5y-2y\)
\(3y=19\)
\(y=19:3\)
\(y=\frac{19}{3}\)
Vậy \(x=12;y=\frac{19}{3}\)
Ta có: \(\frac{9^x}{3^{x+9}}=27\)
\(\Leftrightarrow3^{2x}=3^{x+9}\cdot3^3\)
\(\Leftrightarrow3^{2x}=3^{x+12}\)
\(\Rightarrow2x=x+12\)
\(\Rightarrow x=12\)
Thay vào: \(\frac{4^{12+y}}{2^{5y}}=32\)
\(\Leftrightarrow2^{2y+24}=2^{5y}\cdot2^5\)
\(\Leftrightarrow2^{2y+24}=2^{5y+5}\)
\(\Rightarrow2y+24=5y+5\)
\(\Leftrightarrow3y=19\)
\(\Rightarrow y=\frac{19}{3}\)
Vậy \(\hept{\begin{cases}x=12\\y=\frac{19}{3}\end{cases}}\)