X/12 = Y/8 và x - y = 16
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
áp dung tc cua day ti so bang nhau co
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{-15}{5}=-3\)
x=-6;y=-9
y b lam tuong tuu nhung thay cong bang tru
y c
co \(\frac{x}{y}=\frac{7}{-9}\Rightarrow\frac{x}{7}=\frac{y}{-9}\Rightarrow\frac{2x}{14}=\frac{3y}{-27}\)
lam tuong tuu y a
d,
h cheo
7 ( x + 4 ) = 4 ( 7 + y )
7x + 28 = 4y + 28
7x = 4y
\(\Rightarrow\frac{x}{4}=\frac{y}{7}\)
ap dung tc cua day ti so bang nhau va lam tuong tuu y a
t i c k nha
a) \(|x+7|+|2y-12|=0\)
Vì \(\hept{\begin{cases}|x+7|\ge0;\forall x,y\\|2y-12|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow|x+7|+|2y-12|\ge0;\forall x,y\)
Do đó \(|x+7|+|2y-12|=0\)
\(\Leftrightarrow\hept{\begin{cases}|x+7|=0\\|2y-12|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-7\\y=6\end{cases}}\)
Vậy ...
các phần sau tương tự
a) Ta có :
\(\left|x+7\right|\ge0\)
\(\left|2y-12\right|\ge0\)
Để |x+7| + | 2y - 12| = 0
=> x +7 = 0 và 2y - 12= 0
x = 7 2y = 12
y = 12 : 2
y = 6
Vậy x = 7 ; y = 6
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
\(\left\{{}\begin{matrix}x+y=12\\y+z=-16\\z+x=8\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=12+\left(-16\right)+8\)
\(\Rightarrow x+y+y+z+z+x=4\)
\(\Rightarrow2\left(x+y+z\right)=4\)
\(\Rightarrow x+y+z=2\)
\(\Rightarrow\left\{{}\begin{matrix}z=2-12=-10\\x=2-\left(-16\right)=18\\y=2-8=-6\end{matrix}\right.\)
Từ \(\left\{{}\begin{matrix}x+y=12\\y+z=-16\\x+z=8\end{matrix}\right.\)\(\Rightarrow2\left(x+y+z\right)=4\Rightarrow x+y+z=2\)
*)Xét \(x+y=12\Rightarrow x+y+z=z+12\)
\(\Rightarrow2=z+12\Rightarrow z=-10\)
*)Xét \(y+z=-16\Rightarrow x+y+z=-16+x\)
\(\Rightarrow2=-16+x\Rightarrow x=18\)
*)Xét \(x+z=8\Rightarrow x+y+z=8+y\)
\(\Rightarrow2=8+y\Rightarrow y=6\)
Bài 2:
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Ta có: xy=12
\(\Leftrightarrow12k^2=12\)
\(\Leftrightarrow k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)
a. Ta có: \(\frac{x}{5}=\frac{y}{7}=\frac{x-y}{5-7}=\frac{-12}{-2}=6\)
=> \(\hept{\begin{cases}x=6.5=30\\y=6.7=42\end{cases}}\)
b. x.8 = y. 16
=> \(\frac{x}{16}=\frac{y}{8}=\frac{y-x}{8-16}=\frac{64}{-8}=-8\)
=> \(\hept{\begin{cases}x=-8.16=-128\\y=-8.8=-64\end{cases}}\)
c.Ta có: \(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{x-y}{2+5}=\frac{7}{7}=1\)
=> \(\hept{\begin{cases}x=1.2=2\\y=1.\left(-5\right)=-5\end{cases}}\)
d. Ta có: xy = 10 => x = \(\frac{10}{y}\)(1)
Thay (1) vào \(\frac{x}{2}=\frac{y}{5}\), ta được:
\(\frac{10}{\frac{y}{2}}=\frac{y}{5}\)=> \(\frac{5}{y}=\frac{y}{5}\)
=> y2 = 25
=> y = + 5
y = 5 => x = \(\frac{10}{y}\)= \(\frac{10}{5}\)= 2
y = -5 => x = \(\frac{10}{y}\)= \(\frac{10}{-5}\) = -2
Vậy y = 5; x = 2
y = - 5: x = -2
a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)
Mà \(x-y=-12\)
\(\Rightarrow5k-7k=-12\)
\(\Leftrightarrow-2k=-12\)
\(\Leftrightarrow k=6\)
\(\Rightarrow\hept{\begin{cases}x=5k=30\\y=7k=42\end{cases}}\)
Vậy ...
b) Ta có : \(x.8=y.16\Leftrightarrow\frac{x}{16}=\frac{y}{8}\)
Đặt \(\frac{x}{16}=\frac{y}{8}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=16k\\y=8k\end{cases}}\)
Mà \(y-x=64\)
\(\Rightarrow8k-16k=64\)
\(\Leftrightarrow-8k=64\)
\(\Leftrightarrow k=-2\)
\(\Rightarrow\hept{\begin{cases}x=16k=-32\\y=8k=-16\end{cases}}\)
Vậy ...
\(\frac{x}{12}=\frac{y}{8}\)và \(x-y=16\)
Theo t/c' DTS bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{8}=\frac{x-y}{12-8}=\frac{16}{4}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{12}=4\Rightarrow x=48\\\frac{y}{8}=4\Rightarrow y=32\end{cases}}\)
Vậy x=48,y=32
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{12}=\frac{y}{8}=\frac{x-y}{12-8}=\frac{16}{4}=4\)
\(x=48;y=32\)