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`x-(5/6 -x) =x-2/3`
`x-5/6 +x -x+2/3 =0`
`x = 5/6-2/3 = 5/6 -4/6 = 1/6`
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số số hạng: ( 2500 - 2 ) : 2 +1 = 1250 số
tổng: ( 2500 +2 ) . 1250 : 2 = 1563750
Vì: 1250 . 1251 = 1563750
=> x = 1250
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\(\dfrac{x-6}{50}+\dfrac{x-6}{51}=\dfrac{x-6}{52}+\dfrac{x-6}{53}\)
\(\Rightarrow\dfrac{x-6}{51}+\dfrac{x-6}{50}-\dfrac{x-6}{52}-\dfrac{x-6}{53}=0\)
\(\Rightarrow\left(x-6\right)\left(\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{1}{52}-\dfrac{1}{53}\right)=0\)
\(\Rightarrow x-6=0\) \(\Rightarrow x=6\)
Vậy ...
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x-6/50+x-6/51=x-6/52+x-6/53
x+x-x-x=6/50+6/51-6/52-6/53
0x=6/50+6/51-6/52-6/53(vô ly)
=>ko tồn tại giá trị x
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x^2+2x+6 chia het cho x+4
suy ra x^2+2x+3+2(x+4)+6 chia het cho x+4
x2+2x+6 chia hết cho x+4(vì là bội)
x2+4x-2x-8+14 chia hết cho x+4
x(x+4)-2(x+4)+14 chia hết cho x+4
(x-2)(x+4)+4 chia hết cho x+4
=>4 chia hết cho x+4 hay x+4EƯ(4)={1;-1;2;-2;4;-4}
=>xE{-3;-5;-2;-6;0;-8}
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\(\Leftrightarrow x+3\in\left\{1;-1;3;-3;9;-9\right\}\)
hay \(x\in\left\{-2;-4;0;-6;6;-12\right\}\)
\(\dfrac{x-6}{x+3}=\dfrac{x+3-6}{x+3}=\dfrac{x+3}{x+3}-\dfrac{6}{x+3}=1-\dfrac{6}{x+3}\)
\(\dfrac{x-6}{x+3}⋮x+3\Rightarrow\dfrac{6}{x+3}⋮x+3\\ \Rightarrow x+3\inƯ_{\left(6\right)}=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\)
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X + 1+2+3+4+5-6-7-8-9=1-2-3-4-5+6+7+8+9
X+ (-15) = 17
X = 17-(-15)
X = 32
vậy x = 32
tk nha
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\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{2x^2}{18}=\dfrac{y^2}{36}=\dfrac{2x^2-y^2}{18-36}=\dfrac{-8}{-18}=\dfrac{4}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4.3}{9}=\dfrac{4}{3}\\y=\dfrac{4.6}{9}=\dfrac{8}{3}\end{matrix}\right.\)
Bạn đúng 1 phần, vì đây là 2x2 và y2 nên nó sẽ có 2 trường hợp!
\(\dfrac{x}{3}\)=\(\dfrac{y}{6}\)=\(\dfrac{2x^2}{18}\)=\(\dfrac{y^2}{36}\)=\(\dfrac{2x^2-y^2}{18-36}\)=\(\dfrac{-8}{-18}\) =\(\dfrac{4}{9}\)
=>TH1: \(\dfrac{4}{9}\) ⇒\(\left\{{}\begin{matrix}\dfrac{4}{3}\\\dfrac{8}{3}\end{matrix}\right.\)
=>TH2: \(\dfrac{-4}{9}\)⇒\(\left\{{}\begin{matrix}\dfrac{-4}{3}\\\dfrac{-8}{3}\end{matrix}\right.\)
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Ta có :\(\frac{x}{6}-\frac{7}{y}=\frac{1}{12}\)(y khác 0)
=> \(\frac{xy-42}{6y}=\frac{1}{12}\)
=> 12(xy - 42) = 6y
=> 12xy - 504 = 6y
=> 12xy - 6y = 504
=> 2xy - y = 84
=> y(2x - 1) = 84
Ta có 84 = 1.84 = (-1).(-84) = 42.2 = (-42).(-2) = 21.4 = (-21).(-4) = 7.12 = (-7).(-12) = (-3).(-28) = 28.3 = 14.6 = (-14).(-6)
Lập bảng xét 24 trường hợp
y | 1 | 84 | 42 | 2 | 21 | 4 | 3 | 28 | 6 | 14 | 7 | 12 | -1 | -2 | -3 | -4 | -6 | -7 | -12 | -14 | -21 | -28 | -42 | -84 |
2x - 1 | 84 | 1 | 2 | 42 | 4 | 21 | 28 | 3 | 14 | 6 | 12 | 7 | -84 | -42 | -28 | -21 | -14 | -12 | -7 | -6 | -4 | -3 | -2 | -1 |
x | 42,5 | 1 | 1,5 | 21,5 | 2,5 | 11 | 14,5 | 2 | 7,5 | 3,5 | 6,5 | 4 | -41,5 | -20,5 | -13,5 | -10 | -6,5 | -5,5 | -3 | -2,5 | -1,5 | -1 | -0,5 | 0 |
Vậy các cặp (y;x) thỏa mãn là : (84;1) ; (4 ; 11) ; (12 ; 4) ; (28 ; 2) ; (-4 ; - 10) ; (-12 ; -3) ; (-28 ; -1) ; (-84 ; 0)
x=các số chiha hết cho 6 đều chia gết cho2
x+6 = x+2+4
Ta thay x+2 chia het cho x+2
Nen 4 phai chia het cho x+2 hay x+2 la uoc cua 4
x+2=1;2;4
x=0;2