Tìm x, y ∈ N biết
a) (x - 2)(y + 1) = 14
b) (x + 1)(y + 2) = 20
c) (2x + 1)(y - 2) = 48
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T
1
28 tháng 9 2021
a) \(\left(x-2\right)\left(y+1\right)=14\)
Do \(x,y\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
HH
1
MN
1 tháng 6 2021
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right)\cdot2=-4\\y=\left(-2\right)\cdot5=-10\end{matrix}\right.\)
\(b.\)
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=5\cdot4=20\end{matrix}\right.\)
17 tháng 3 2023
a: 3x=7y
=>x/7=y/3=(x-y)/(7-3)=-16/4=-4
=>x=-28; y=-12
b: x/6=y/5
=>x/6=2y/10=(x+2y)/(6+10)=20/16=5/4
=>x=30/4=15/2; y=25/4
c: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot2+3\cdot\left(-3\right)+5\cdot5}=\dfrac{6}{20}=\dfrac{3}{10}\)
=>x=3/5; y=-9/10; z=3/2
d: x/2=y/3
=>x/8=y/12
y/4=z/5
=>y/12=z/15
=>x/8=y/12=z/15
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
=>x=16; y=24; z=30
N
1
9 tháng 12 2021
\(a,\text{Vì }x,y\in N\Leftrightarrow x+2\ge2;y+3\ge3\\ \Leftrightarrow\left(x+2\right)\left(y+3\right)=6=2\cdot3=3\cdot2\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;0\right)\)
\(b,\Leftrightarrow\left(x-3\right)\left(y+1\right)=7\cdot1=1\cdot7\\ \left\{{}\begin{matrix}x-3=7\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x-3=1\\y+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;0\right);\left(4;6\right)\right\}\)
16 tháng 10 2021
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
NN
1
4 tháng 2 2021
a) Ta có: (x+1)(y-2)=-2
nên x+1; y-2 là các ước của -2
Trường hợp 1:
\(\left\{{}\begin{matrix}x+1=-1\\y-2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=4\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x+1=2\\y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x+1=-2\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=3\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x+1=1\\y-2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy: (x,y)\(\in\){(-2;4);(1;1);(-3;3);(0;0)}
b) Ta có: (x+1)(xy-1)=3
nên x+1;xy-1 là các ước của 3
Trường hợp 1:
\(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\Leftrightarrow loại\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-\dfrac{1}{2}\end{matrix}\right.\left(loại\right)\)
Vậy: \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;1\right)\right\}\)
c) Ta có: \(\left(x+y\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vây: (x,y)=(-1;1)
d) Ta có: \(\left|x+y\right|\cdot\left(x-y\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+y\right|=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy: (x,y)=(0;0)
TV
1
AH
Akai Haruma
Giáo viên
30 tháng 10 2021
Lời giải:
a. Đề có cả x,y. Bạn xem lại
b.
PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$
$\Leftrightarrow (x-3)(5x-2)=0$
$\Leftrightarrow x-3=0$ hoặc $5x-2=0$
$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$
c.
PT $\Leftrightarrow (7x-2)(x-4)=0$
$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$
$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$
d. Đề thiếu.
3 tháng 7 2021
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!
a) x=3 y=13
x=16 y=0
x=4 y=5
x=9 y=1