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Ta có:
\(\frac{27^x}{3^{2x-y}}=243=3^5\Rightarrow27^x=3^5.3^{2x-y}=3^{5+2x-y}\Rightarrow3^{3x}=3^{5+2x-y}\Rightarrow3x=5+2x-y\Rightarrow3x-2x=5-y\Rightarrow x=5-y\)(1)\(\frac{25^x}{5^{x+y}}=125=5^3\Rightarrow25^x=5^3.5^{x+y}\Rightarrow5^{2x}=5^{3+x+y}\Rightarrow2x=3+x+y\Rightarrow2x-x=3+y\Rightarrow x=3+y\)(2)
Từ (1) và (2)⇒
\(x=5-y=3+y\Rightarrow y=1\Rightarrow x=4\)
Vậy y=1; x=4 thỏa mãn đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}3^{3x-2x+y}=3^5\\5^{2x-x-y}=5^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\x-y=3\end{matrix}\right.\)
=>x=1;y=-2
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Bài 1:
Bài 2:
\(\frac{4^x}{2^{x+y}}=8\Leftrightarrow4^x=8.2^{x+y}\Leftrightarrow\left(2^2\right)^x=2^3.2^{x+y}\Leftrightarrow2^{2x}=2^{x+y+3}\)<=>2x=x+y+3<=>x=y+3
\(\frac{9^{x+y}}{3^{5y}}=243\Leftrightarrow9^{x+y}=243.3^{5y}\Leftrightarrow\left(3^2\right)^{x+y}=3^5.3^{5y}\Leftrightarrow3^{2x+2y}=3^{5y+5}\)<=>2x+2y=5y+5
<=>2x=3y+5 mà x=y+3 => 2(y+3)=3y+5 <=> 2y+6=3y+5 <=> 6-5=3y-2y <=> y=1 <=> x=1+3=4
Vậy xy=4.1=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{x+2}+5^{x+3}=750\)
\(5^x.5^2+5^x.5^3=750\)
\(5^x.25+5^x\cdot125=750\)
\(5^x.\left(25+125\right)=750\)
\(5^x.150=750\)
\(5^x=750:150\)
\(5^x=5\)
\(5^x=5^1\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x^3}{27}=\frac{y^3}{64}=\frac{z^3}{125}=>\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=>\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{3z^2}{75}\)
áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{3z^2}{75}=\frac{x^2+y^2-3z^2}{9+16-75}=\frac{50}{-50}=-1\)
\(\frac{x}{3}=-1=>x=-3\)
\(\frac{y}{4}=-1=>y=-4\)
\(\frac{z}{5}=-1=>z=-5\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
\(\frac{x}{10}=2\Rightarrow x=10.2=20\)
\(\frac{y}{6}=2\Rightarrow y=2.6=12\)
\(\frac{z}{21}=2\Rightarrow z=21.2=42\)
\(d,\frac{x}{2}=\frac{y}{3}=k\)\(\Rightarrow x=2k;y=3k\)
\(\Rightarrow ab=2k.3k=6k^2=54\)
\(\Rightarrow k^2=9\Leftrightarrow k=3\)
\(\frac{x}{2}=3\Rightarrow x=6\)
\(\frac{y}{3}=3\Rightarrow y=9\)
a) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\) => \(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
=> \(\hept{\begin{cases}\frac{x}{10}=2\\\frac{y}{6}=2\\\frac{z}{21}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.10=20\\y=2.6=12\\z=2.21=42\end{cases}}\)
Vậy x = 20; y = 12; z = 42
b) Ta có: \(\frac{x}{3}=\frac{y}{4}\) => \(\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{7}\) => \(\frac{y}{20}=\frac{z}{28}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)=> \(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{125}{62}=\frac{125}{62}\)
=> \(\hept{\begin{cases}\frac{x}{15}=\frac{125}{62}\\\frac{y}{20}=\frac{125}{62}\\\frac{z}{28}=\frac{125}{62}\end{cases}}\) => \(\hept{\begin{cases}x=\frac{125}{62}.15=\frac{1875}{62}\\y=\frac{125}{62}.20=\frac{1250}{31}\\z=\frac{125}{62}.28=\frac{1750}{31}\end{cases}}\)
Vậy ...
minh bik lam bai b thui a
25^x=5^2.x:5^1.x+y=5^3
suy ra 2.x : 1.x+y=3