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a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)
áp dụng tính chất dãy tỉ số = nhau
\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)
\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)
\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)
\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)
b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)
có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)
\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)
áp dụng t/c dãy tỉ số = nhau
\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)
\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)
\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)
\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)
c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)
thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(=>y=2\)
\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)
d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)
thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)
\(=>x=\dfrac{2.3}{3}=2\)
c, từ đoạn này á
\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)
\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a )
Ta có :
\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{20}=\frac{y}{24}\\\frac{y}{24}=\frac{z}{21}\end{cases}}}\)
và \(x+y-z=69\)
ADTCDTSBN , ta có :
\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{20}=3\\\frac{y}{24}=3\\\frac{z}{21}=3\end{cases}\Rightarrow\hept{\begin{cases}x=3.20=60\\y=3.24=72\\z=3.21=63\end{cases}}}\)
Vậy ...
b )
Ta có :
\(5y=72\Rightarrow y=\frac{72}{5}=14,4\)
\(\Rightarrow x=14,4.3:2=21,6\)
và \(3x+5y-7z=30\)
Thay vào làm tiếp :
c )
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)
\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)
\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)
\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)( ADTCDTSBN )
\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=2\\\frac{y+3}{4}=2\\\frac{z-5}{6}=2\end{cases}\Rightarrow\hept{\begin{cases}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=5\\z=17\end{cases}}}\)
Vậy ...
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405
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suy ra:x/2=y/3 , y/7 = z/5
suy ra x/14 = y/21 = z/15 = 3x/42 = 5y/105 = 7z/105
áp dụng tính chất của dãy tỉ số bằng nhau ta có :
3x/42= 5y/105 = 7z/105= 3x +5y -7z/42+105-105 = 10/7
suy ra : x= 20
y = 30
z = 150/7
Néu đúng thì k cho mk nha
ta có: 3x=2y => \(\frac{x}{2}=\frac{y}{3}\)
5y=7z =>\(\frac{y}{7}=\frac{z}{5}\)
=>\(\frac{x}{14}=\frac{y}{21}=\frac{z}{15}\)
=>\(\frac{3x}{42}=\frac{5y}{105}=\frac{7z}{245}=\)\(\frac{3x+5y-7z}{42+105-105}=\frac{60}{42}=\frac{10}{7}\)
\(\frac{x}{14}=\frac{10}{7}\)=> x =20
\(\frac{y}{21}=\frac{10}{7}\)=> y = 30
\(\frac{z}{15}=\frac{10}{7}\) => z=\(\frac{150}{7}\)
Đáp số:20;30;\(\frac{150}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có: \(2x=3y;5y=7z\)và \(x-y-z=-27\)
\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)và\(x-y-z=-27\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)và \(x-y-z=-27\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y-z}{21-14-10}=\frac{-27}{-3}=9\)
Ta có:\(\frac{x}{21}=9\Rightarrow x=9.21=189\)
\(\frac{y}{14}=9\Rightarrow y=9.14=126\)
\(\frac{z}{10}=9\Rightarrow z=9.10=90\)
Vậy:\(x=189;y=126\)và\(z=90\)
b) \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)và\(x^2-2y^2+z^2=18\)
\(\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)và\(x^2-2y^2+z^2=18\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}=\frac{x^2-2y^2+z^2}{16-50+36}=\frac{18}{2}=9\)
Ta có:\(\frac{x^2}{16}=9\Rightarrow x^2=144\Rightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
\(\frac{2y^2}{50}=9\Rightarrow2y^2=450\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)
\(\frac{z^2}{36}=9\Rightarrow z^2=324\Rightarrow\orbr{\begin{cases}z=18\\z=-18\end{cases}}\)
Vậy: \(x=12;y=15;z=18\)hoặc \(x=-12;y=-15;z=-18\)
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\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2};5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}=\frac{7y}{14};\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{2y}{14}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{5y}{70}=\frac{7z}{70}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3x}{63}=\frac{5y}{70}=\frac{7z}{70}=\frac{3x+5y-7z}{63+70-70}=\frac{30}{63}=\frac{10}{21}\)
\(\frac{3x}{63}=\frac{10}{21}\Rightarrow x=\frac{10}{21}.63:3=10\)
\(\frac{5y}{70}=\frac{10}{21}\Rightarrow y=\frac{10}{21}.70:5=\frac{20}{3}\)
\(\frac{7z}{70}=\frac{10}{21}\Rightarrow z=\frac{10}{21}.70:7=\frac{100}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
có 3x=5y=7z
\(\Rightarrow\frac{x}{35}=\frac{y}{21}=\frac{x}{15}\) (z/15 nha, ko phải x/15)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}=\frac{x+y-z}{35+21-15}=\frac{41}{41}=1\)
=>\(\frac{x}{35}=1\Rightarrow x=35\)
\(\frac{y}{21}=1\Rightarrow y=21\)
\(\frac{z}{15}=1\Rightarrow z=15\)
vậy...........
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$
$\Rightarrow \frac{x}{21}=\frac{y}{14}$
$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}$
Vậy:
$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$
$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$
$\Rightarrow x=21.2=42; y=14.2=28; z=10.2=20$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
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 \((x,y)\in(6,15);(-6,-15)\)
\(3x=5y=7z\)
=> \(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}=\frac{x+y-z}{35+21-15}=\frac{41}{41}=1\)
đến đây tự tính nhé
Đặt : \(3x=5y=7x=k\)
\(\Rightarrow x=\frac{k}{3};\)\(y=\frac{k}{5};\)\(z=\frac{k}{7}\)
\(\Rightarrow x+y-z=\frac{k}{3}+\frac{k}{5}-\frac{k}{7}=\frac{41k}{105}=41\)
\(\Rightarrow41k=41.105\)
\(\Rightarrow k=105\)
\(\Rightarrow x=\frac{105}{3}=35;\)\(y=\frac{105}{5}=21;\)\(z=\frac{105}{7}=15\)
----------------------HOK TỐT---------------------------