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Gọi số cây trồng của ba lớp `7A,7B,7C` lần lượt là `x,y,z` (cây; `x,y,z>0`)
`-` Theo đề bài, ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) và `x+y+z=120`
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{3+4+5}=\dfrac{120}{12}=12\)
\(\Rightarrow\\ x=3\cdot10=30\\ y=4\cdot10=40\\ z=5\cdot10=50\)
Vậy, số cây trồng được của ba lớp `7A,7B,7C` lần lượt là `30,40,50` cây.
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,` Áp dụng t/c dãy tỉ số bằng nhau ta có :
`x/2=y/3=z/4 = (x+y+z)/(2+3+4)=-18/9 =-2`
`=> x/2=-2=>x=-2*2=-4`
`=>y/3=-2=>y=-2*3=-6`
`=>z/4=-2=>z=-2*4=-28`
`b,` Áp dụng t/c dãy tỉ số bằng nhau ta có :
`x/5 =y/6=z/7 = (x-y+z)/(5-6+7)=36/6=6`
`=>x/5=6=>x=6*5=30`
`=>y/6=6=>y=6*6=36`
`=>z/7=6=>z=6*7=42`
`@ Kidd`
![](https://rs.olm.vn/images/avt/0.png?1311)
M(x)=x^2-6x+15
=x^2-6x+9+6
=(x-3)^2+6>=6>0 với mọi x
=>M(x) ko có nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
i: \(=\dfrac{15}{34}+\dfrac{19}{34}+\dfrac{7}{21}+\dfrac{2}{3}-1-\dfrac{15}{37}\)
\(=1+\dfrac{1}{3}+\dfrac{2}{3}-1-\dfrac{15}{37}\)
\(=1-\dfrac{15}{37}=\dfrac{22}{37}\)
j: \(=1-\left(-27\right)+\dfrac{1}{2}:\dfrac{-1}{8}\)
\(=1+27-4=24\)
k: \(=-8+\dfrac{1}{2}\cdot\dfrac{-4}{1}-15\)
\(=-8-2-15=-25\)
l: \(=3:\dfrac{9}{4}+\dfrac{1}{9}\cdot6\)
\(=3\cdot\dfrac{4}{9}+\dfrac{1}{9}\cdot6\)
\(=\dfrac{4}{3}+\dfrac{2}{3}=2\)
m: \(=9\cdot\dfrac{1}{3}-\left(-27\right)=3+27=30\)
n: \(\sqrt{\dfrac{16}{25}}\cdot\sqrt{\dfrac{121}{64}}-1\dfrac{3}{10}\)
\(=\dfrac{4}{5}\cdot\dfrac{11}{8}-\dfrac{13}{10}\)
\(=\dfrac{11}{10}-\dfrac{13}{10}=-\dfrac{2}{10}=-\dfrac{1}{5}\)
o: \(=\dfrac{9}{8}\cdot12-\dfrac{1}{2}\)
\(=\dfrac{27}{2}-\dfrac{1}{2}=\dfrac{26}{2}=13\)
p: \(=\dfrac{3^2\cdot2^2+3^2\cdot3\cdot2^2+3^2}{-13}\)
\(=\dfrac{3^2\left(2^2+3\cdot2^2+3^2\right)}{-13}\)
\(=\dfrac{9\cdot\left(4+3\cdot4+9\right)}{-13}\)
\(=\dfrac{9\cdot25}{-13}=-\dfrac{225}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
2x=3y = \(\dfrac{x}{3}\) = \(\dfrac{y}{2}\) = \(\dfrac{x}{15}\) = \(\dfrac{y}{10}\)
3y=5z = \(\dfrac{y}{5}\) = \(\dfrac{z}{3}\) = \(\dfrac{y}{10}\) = \(\dfrac{z}{6}\)
=> \(\dfrac{x}{15}\) = \(\dfrac{y}{10}\) = \(\dfrac{z}{6}\) = \(\dfrac{x.y.z}{15.10.6}\) = \(\dfrac{36}{900}\)= \(\dfrac{1}{25}\)
=> x= \(\dfrac{1}{25}\) . 15 =\(\dfrac{3}{5}\)
y=\(\dfrac{1}{25}\) . 10 = \(\dfrac{2}{5}\)
z=\(\dfrac{1}{25}\).6 = \(\dfrac{6}{25}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
1/2-(2/3.x-1/3)=2/3
1/2-2/3x+1/3=2/3
1/2-2/3x=2/3-1/3
1/2-2/3x=1/3
2/3x=1/2-1/3
2/3x=1/6
x=1/6:2/3
x=1/4
vậy x bằng 1/4