3 / 5 x y = 4 / 7
25 / 3 - y = 5/6
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\(\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{2}{4}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{6}{12}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\dfrac{7}{12}\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{36}{60}\right)-\dfrac{35}{60}\right]\\ =\dfrac{3}{4}-\left(-\dfrac{71}{60}\right)\\ =\dfrac{3}{4}+\dfrac{71}{60}\\ =\dfrac{35}{60}+\dfrac{71}{60}\\ =\dfrac{106}{60}\\ =\dfrac{53}{30}\)
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\(-\dfrac{6}{7}.\dfrac{21}{12}\\ =-\dfrac{126}{84}\\ =-\dfrac{63}{42}\\ =-\dfrac{31}{14}\)
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\(\left(-5\right).\left(-\dfrac{6}{20}\right)\\ =\dfrac{\left(-5\right).\left(-6\right)}{20}\\ =\dfrac{30}{20}\\ =\dfrac{3}{2}\)
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Gọi x,y là nghiệm của phương trình:
\(\left\{{}\begin{matrix}S=x+y=3\\P=x.y=2\end{matrix}\right.\Rightarrow a^2-S.a+P=0\)
\(\Leftrightarrow a^2-3a+2=0\Leftrightarrow\left[{}\begin{matrix}a_1=x=2\\a_2=y=1\end{matrix}\right.\)
a)\(x^2+y^2=1^2+2^2=5\)
b)\(x^3+y^3=1^3+2^3=9\)
c)\(x^4+y^4=1^4+2^4=17\)
d)\(x^5+y^5=1^5+2^5=33\)
e)\(x^6+y^6=1^6+2^6=65\)
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
2:
a: 5/x-y/3=1/6
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)
=>30-2xy=x
=>x(2y+1)=30
=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}
=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}
b: x/6-2/y=1/30
=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)
=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)
=>5xy-60=y
=>y(5x-1)=60
=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)
=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}
1)\(\left(x+1\right).\left(y-2\right)=0\) \(\left(x,y\inℤ\right)\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
2)\(\left(x-5\right).\left(y-7\right)=1\)
x-5 | 1 | -1 |
y-7 | 1 | -1 |
x | 6 | 4 |
y | 8 | 6 |
3)\(\left(x+4\right).\left(y-2\right)=2\)
x+4 | 1 | 2 | -1 | -2 |
y-2 | 2 | 1 | -2 | -1 |
x | -3 | -2 | -5 | -6 |
y | 4 | 3 | 0 | 1 |
4)\(\left(x-4\right).\left(y+3\right)=-3\)
x-4 | 1 | -1 | 3 | -3 |
y+3 | -3 | 3 | -1 | 1 |
x | 5 | 3 | 7 | 1 |
y | -6 | 0 | -4 | -2 |
5)\(\left(x+3\right).\left(y-6\right)=-4\)
x+3 | -1 | 1 | -4 | 4 | 2 | -2 |
y-6 | 4 | -4 | 1 | -1 | -2 | 2 |
x | -4 | -2 | -7 | 1 | -1 | -5 |
y | 10 | 2 | 7 | 5 | 4 | 8 |
6)\(\left(x-8\right).\left(y+7\right)=5\)
x-8 | 1 | 5 | -1 | -5 |
y+7 | 5 | 1 | -5 | -1 |
x | 9 | 13 | 7 | 3 |
y | -2 | -6 | -12 | -8 |
7)\(\left(x+7\right).\left(y-3\right)=-6\)
x+7 | -1 | 1 | -6 | 6 | -2 | 2 | -3 | 3 |
y-3 | 6 | -6 | 1 | -1 | 3 | -3 | 2 | -2 |
x | -8 | -6 | -13 | -1 | -9 | -5 | -10 | -4 |
y | 9 | -3 | 4 | 2 | 6 | 0 | 5 | 1 |
8)\(\left(x-6\right).\left(y+2\right)=7\)
x-6 | 1 | 7 | -1 | -7 |
y+2 | 7 | 1 | -7 | -1 |
x | 7 | 13 | 5 | -1 |
y | 5 | -1 | -9 | -3 |
ok :)
y = 4/7 :3/5
y = 20/21
y= 25/3- 5/6
y= 15/2
\(\dfrac{3}{5}\times y=\dfrac{4}{7}\)
\(y=\dfrac{4}{7}\div\dfrac{3}{5}\)
\(y=\dfrac{4}{7}\times\dfrac{5}{3}\)
\(y=\dfrac{20}{21}\)