3.(x+y)-3y(x+y) cứu mình với :(
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=(3x+3y)-(x^2+2xy+y^2)=3(x+y)-(x+y)^2=k rõ nữa
=(4x^2-4xy) -(6y^2-6xy)= 4x(x-y)+6y(x-y)=2(x-y)(2x+3y)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)
a) \(\frac{x}{y}=\frac{5}{7}\)=>\(\frac{x}{5}=\frac{y}{7}=>\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\frac{xy}{5.7}\)
=>\(\frac{x^2}{25}=\frac{y^2}{49}=\frac{35}{35}=1\)
=> \(x^2=25;y^2=49\)
=>\(x=\pm5;y=\pm7\)
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hc tốt
\(3x=2y=z\Rightarrow\frac{z}{6}=\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\frac{z}{6}=\frac{x}{2}=\frac{y}{3}=\frac{x+y+z}{6+2+3}=\frac{99}{11}=9\)
\(\Rightarrow\hept{\begin{cases}z=54\\x=18\\y=27\end{cases}}\)
Bạn chắc đây là Toán lớp 4 không ạ?
\(3\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\cdot\left(3-3y\right)\)
\(=\left(x+y\right)\left(3\cdot1-3\cdot y\right)\)
\(=3\cdot\left(x+y\right)\cdot\left(1-y\right)\)