6/4 - 8/10 - y=4/6 . Tìm y?
E+x^(4)*y^(4)+x^(5)*y^(5)+x^(6)*y^(6)+x^(7)*y^(7)+x^(8)*y^(8)+x^(9)*y^(9)+x^(10)*y^(10) tại x=-1, y=1
E = x^(4)*y^(4)+x^(5)*y^(5)+x^(6)*y^(6)+x^(7)*y^(7)+x^(8)*y^(8)+x^(9)*y^(9)+x^(10)*y^(10) tại x=-1, y=1 nha
Tim y biet : ( 1/2*4 + 1/4*6 + 1/6*8+ 1/8*10 ) * y = 1/3
Tìm cặp số nguyên x, y biết :
a) (x + 4).( y – 8) = 6
b) 2x + xy + 3y + 6 = 10
a) x,y nguyên => x+4; y-8 nguyên
=> x+4; y-8\(\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
ta có bảng
x+4 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
x | -10 | -7 | -6 | -5 | -3 | -2 | -1 | 2 |
y-8 | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
y | 7 | 6 | 5 | 2 | 14 | 11 | 10 | 9 |
Vậy (x;y)={(-10;7);(-7;6);(-6;5);(-5;2);(-3;14);(-2;11);(-1;10);(2;9)}
b) 2x+xy+3y+6=10
<=> x(2+y)+3(y+2)=10
<=> (y+2)(x+3)=10
x,y nguyên => y+2; x+3 nguyên
=> y+2; x+3\(\in\)Ư(10)={-10;-5;-2;-1;1;2;5;10}
ta có bảng
x+3 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
x | -13 | -8 | -5 | -4 | -2 | -1 | 2 | 7 |
y+2 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
y | -3 | -4 | -7 | -12 | 8 | 3 | 0 | -1 |
Tìm y
a) ( 1/2*4 + 1/4*6 + 1/6*8 + 1/8*10) * y =1/3
b) ( 1/1*3 + 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11) * y =2/3
Help me please!!!!!!!!!!!!!
\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\frac{2}{5}.y=\frac{1}{3}\)
\(y=\frac{1}{3}:\frac{2}{5}\)
\(y=\frac{5}{6}\)
\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}:\frac{10}{11}\)
\(y=\frac{22}{30}\)
Tìm x,y,z thuộc z, biết:
x/-10 = -7/y =2/-6=-4/8
TÌM Y
1) [( 3/8*y-1/2): 5/7+4/9 ]*6/11=18/55
2) ( 7/10-3*y):(1 9/10-1 2/5)+4/5=1
Tìm cặp số nguyên x, y biết :
a) (x + 4).( y – 8) = 6
b) 2x + xy + 3y + 6 = 10
tìm các số nguyên x,y,z a) -4/8 = x/-10 = -7/y = z/-24 b) -3/6 = x/-2 = -18/y = -z/24
a: \(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\)
=>\(\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\left\{{}\begin{matrix}x=\left(-10\right)\cdot\dfrac{\left(-1\right)}{2}=5\\y=\dfrac{-7\cdot2}{-1}=14\\z=\dfrac{-24\cdot\left(-1\right)}{2}=\dfrac{24}{2}=12\end{matrix}\right.\)
b: \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
=>\(\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\dfrac{x}{2}=\dfrac{18}{y}=\dfrac{z}{24}=\dfrac{1}{2}\)
=>\(x=2\cdot\dfrac{1}{2}=1;y=18\cdot\dfrac{2}{1}=36;z=\dfrac{24}{2}=12\)
Tìm y
c } y + 6 y 8 - 4 = 60
d } { y + 6 } x 8- 4 = 60