Question

[Hoàng Nhật Thiên Vân]

Tìm x; y biết:

a) \(\dfrac{x}{3}\) = \(\dfrac{y}{12}\) và x + y = 5

b) x : 2 = y : (-5) và x - y = -14

c) 13x = 7x và x + y = 40

d) \(\dfrac{4}{x}\) = \(\dfrac{5}{y}\) và x + y = 36

Answer 1

[Nguyễn Thanh Hằng]

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{3}=\dfrac{y}{12}=\dfrac{x+y}{3+12}=\dfrac{5}{15}=\dfrac{1}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{1}{3}\\\dfrac{y}{12}=\dfrac{1}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

Vậy ...

b, \(x:2=y:\left(-5\right)\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Theo t.c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-14}{7}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-2\\\dfrac{y}{-5}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=10\end{matrix}\right.\)

Vậy ...

c, \(13x=7x\Leftrightarrow\dfrac{13x}{91}=\dfrac{7x}{91}\)

\(\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{13}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=2\\\dfrac{y}{13}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=14\\y=26\end{matrix}\right.\)

Vậy ....

d, \(\dfrac{4}{x}=\dfrac{5}{y}\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{5}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{36}{9}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=4\\\dfrac{y}{5}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=16\\y=20\end{matrix}\right.\)

Vậy ..

Answer 2

[Mysterious Person]

a) ta có : \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{12}\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{12}=0\\x+y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-y=0\\x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\) vậy \(x=1;y=4\)

b) ta có : \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{-5}\\x-y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{5}=0\\x-y=-14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x+2y=0\\2x-2y=-28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=-28\\x-y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=10\end{matrix}\right.\) vậy \(x=-4;y=10\)

c) ta có : \(\left\{{}\begin{matrix}13x=7y\\x+y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x-7y=0\\x+y=40\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}13x-7y=0\\7x+7y=280\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20x=280\\x+y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=14\\y=26\end{matrix}\right.\) vậy \(x=14;y=26\)

) ta có : \(\left\{{}\begin{matrix}\dfrac{4}{x}=\dfrac{5}{y}\\x+y=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=4y\\x+y=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=0\\4x+4y=144\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9x=144\\x+y=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\y=20\end{matrix}\right.\) vậy \(x=16;y=20\)