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11 tháng 10 2023
b:
ĐKXĐ: x<>0
\(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)
=>\(6\left(6+xy\right)=3x\)
=>\(x=2\left(6+xy\right)=12+2xy\)
=>\(x\left(1-2y\right)=12\)
mà x,y là các số nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)
c: ĐKXĐ: y<>-1
\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)
=>\(2xy+2x+6=y+1\)
=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)
=>\(\left(2x-1\right)\left(y+1\right)=-6\)
mà x,y là các số nguyên
nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)
3 tháng 3 2022
a:\(A=5:\dfrac{1}{2}+\dfrac{20}{5}+1:\dfrac{-1}{4}=10+4-4=10\)
b: y/x=1/4
nên x=4y
\(A=\dfrac{4x+7y}{x-3y}=\dfrac{16y+7y}{4y-3y}=23\)
11 tháng 10 2023
d:
ĐKXĐ: y<>0; x<>0; y<>2
\(\dfrac{4}{x}+\dfrac{2}{y}=1\)
=>\(\dfrac{4y}{xy}+\dfrac{2x}{xy}=1\)
=>2x+4y=xy
=>x(2-y)=-4y
=>x(y-2)=4y
=>\(x=\dfrac{4y}{y-2}\)
mà x,y nguyên
nên \(4y⋮y-2\)
\(\Leftrightarrow4y-8+8⋮y-2\)
=>\(y-2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(y\in\left\{3;1;4;6;-2;10;-6\right\}\)
=>\(x\in\left\{12;-4;8;6;2;5;3\right\}\)
e:
ĐKXĐ: x<>0; y<>0; y<>3
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\)
=>\(\dfrac{x+y}{xy}=\dfrac{1}{3}\)
=>3x+3y=xy
=>x(3-y)=-3y
=>\(x=\dfrac{3y}{y-3}\)
mà x,y nguyên
nên \(3y⋮y-3\)
=>\(3y-9+9⋮y-3\)
=>\(y-3\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(y\in\left\{4;2;6;12;-6\right\}\)
=>\(x\in\left\{12;-6;6;4;2\right\}\)
AH
Akai Haruma
Giáo viên
17 tháng 12 2021
Lời giải:
Nếu $x+y+z=0$ thì:
$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$
$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$
$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$
(thỏa mãn đkđb)
Khi đó:
$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$
$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$
Nếu $x+y+z\neq 0$
Áp dụng TCDTSBN:
$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$
$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:
$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$
22 tháng 10 2017
a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x
Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y
= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y
Ta lại có : 1 + 4y/24 = 1+4y / 9+3y
=> 24=9+3y => 15=3y => y=5
Vậy y=5
Nhớ like
22 tháng 10 2017
b, 1+3y/12 = 1+5y/5x = 1+7y/4x
Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x
= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x
Ta lại có: 1+5y / 5x = 1+5y / 6+2x
=> 5x = 6+2x => 3x = 6 => x=2
Vậy x =2
AH
Akai Haruma
Giáo viên
16 tháng 1 2018
Lời giải:
\(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(A+3=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{x+y}+1\right)\)
\(A+3=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\)
\(A+3=2017\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(A+3=2017.\frac{1}{672}=\frac{2017}{672}\)
\(\Rightarrow A=\frac{2017}{672}-3=\frac{1}{672}\)
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{617}\)
\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{617}\)
\(\Rightarrow617\left(x+y\right)=xy\)
\(\Rightarrow617x+617y=xy\)
\(\Rightarrow617x+617y-xy+617^2=617^2\)
\(\Rightarrow\left(y-617\right)\left(x-617\right)=617^2\)
\(\Rightarrow\left[{}\begin{matrix}x-617=617\\y-617=617\end{matrix}\right.\)
\(\Rightarrow x=y=1234\)