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\(3x^2y^3-x^2y-M=x^2y^3+x^2y\\ \Rightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y\\ \Rightarrow M=2x^2y^3-2x^2y\)
\(\Leftrightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y=2x^2y^3-2x^2y\)
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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
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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 \(\hept{\begin{cases}x=10.\frac{2}{3}=\frac{20}{3}\\y=15.\frac{2}{3}=10\\z=21.\frac{2}{3}=14\end{cases}}\)
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c) 3x + 4 + 3x + 2 = 810
=> 3x . 34 + 3x . 32 = 810
=> 3x.(34 + 32) = 810
=> 3x . (81 + 9) = 810
=> 3x . 90 = 810
=> 3x = 810 : 90
=> 3x = 9
=> 3x = 32
=> x = 2
d) 3x + 3x + 2 = 810
=> 3x + 3x . 32 = 810
=> 3x . (1 + 32) = 810
=> 3x . (1 + 9) = 810
=> 3x . 10 = 810
=> 3x = 810 : 10
=> 3x = 81
=> 3x = 34
=> x = 4
c, \(3^{x+4}+3^{x+2}=810\)
\(\Leftrightarrow3^x\left(3^4+3^2\right)=810\)
\(\Leftrightarrow3^x.90=810\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
d, \(3^x+3^{x+2}=810\)
\(\Leftrightarrow3^x\left(1+3^2\right)=810\)
\(\Leftrightarrow3^x.10=810\)
\(\Leftrightarrow3^x=81=3^4\)
\(\Leftrightarrow x=4\)
P/s: Toán thường thôi nhỉ :) Ko nâng cao lắm
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\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chát dãy tỉ số = nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=3\Rightarrow z=63\)
b, Tự làm
c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)
\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)
\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)
\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)
Vậy \((x,y)\in(6,15);(-6,-15)\)
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`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
\(3^x+3^{x-2}=810\)
<=> \(3^x+3^x:3^2=810\)
<=> \(3^x\left(1+\frac{1}{9}\right)=810\)
<=> \(3^x=729\)
<=>\(3^x=3^6\)
=> x=6
3x +3x-2 = 810
\(\Rightarrow3^x\left(1+3^{-2}\right)=810\)
\(\Rightarrow3^x\left(1+\frac{1}{3^2}\right)=810\)
\(\Rightarrow3^x\left(1+\frac{1}{9}\right)=810\)
\(\Rightarrow3^x\cdot\frac{10}{9}=810\)
\(\Rightarrow3^x=729\)
\(\Rightarrow3^x=3^6\Leftrightarrow x=6\)