(3*y)*2-7=29
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NT
1
31 tháng 10 2016
Ta có (x - y)2 = x2 - 2xy + y2 = 29 - 2xy = 49
=> xy = - 10
Ta lại có
x3 - y3 = (x - y)(x2 + xy + y2) = 7(29 - 10) = 133
4 tháng 11 2021
b: \(=\dfrac{2}{7}-\dfrac{3}{7}\cdot\dfrac{-2}{3}=\dfrac{2}{7}+\dfrac{2}{7}=\dfrac{4}{7}\)
c: \(=\dfrac{3}{7}-\dfrac{7}{2}-\dfrac{3}{7}+\dfrac{7}{2}=0\)
NB
0
LD
1
5 tháng 12 2022
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{5}=\dfrac{y-1}{7}=\dfrac{z}{3}=\dfrac{2x+y-z-1}{5\cdot2+7-3}=\dfrac{28}{14}=2\)
=>x=10; y-1=14; z=6
=>x=10; y=15; z=6
12 tháng 3 2022
a: =(5/7+2/7)+(4/3+5/3)=3+1=4
b: =(17/12+7/12)+(29/7-8/7)
=2+3=5
c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)
=1+2+1
=4
d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)
=1+1+3=5
MA
2
10 tháng 10 2020
a) Ta có: \(x\left(y-7\right)+y-7=41\)
\(\Leftrightarrow\left(y-7\right)\left(x+1\right)=41\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-7=41\\x+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=-41\\x+1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=1\\x+1=41\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=-1\\x+1=-41\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=48\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=-34\\x=-2\end{matrix}\right.\\\left\{{}\begin{matrix}y=8\\x=40\end{matrix}\right.\\\left\{{}\begin{matrix}y=6\\x=-42\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(0;48);(-2;-34);(40;8);(-42;6)}
b) Ta có: \(x\left(y-3\right)-y+3=29\)
\(\Leftrightarrow x\left(y-3\right)-\left(y-3\right)=29\)
\(\Leftrightarrow\left(y-3\right)\left(x-1\right)=29\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-3=29\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=-29\\x-1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=1\\x-1=29\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=-1\\x-1=-29\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=32\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-26\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=4\\x=30\end{matrix}\right.\\\left\{{}\begin{matrix}y=2\\x=-28\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(2;32);(0;-26);(30;4);(-28;2)}
TN
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 12 2022
a, 3x ( y+1) + y + 1 = 7
(y+1)(3x +1) =7
th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)
th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)
th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)
Vậy (x,y)= (2 ;0); (0; 6)
b, xy - x + 3y - 3 = 5
(x( y-1) + 3( y-1) = 5
(y-1)(x+3) = 5
th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)
th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)
th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)
vậy (x, y) = ( 8; 2); ( -8; 0); (-2; 6); (-4; -4)
c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1
⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1 ⋮ 2x + 1
th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8
th2: 2x+ 1 = 1=> x =0; y = 7
th3: 2x+1 = -3 => x = x=-2 => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3
th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2
th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2
th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1
th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1
th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0
kết luận
(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)
26 tháng 12 2022
3xy−2x+5y=293xy−2x+5y=29
9xy−6x+15y=879xy−6x+15y=87
(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77
3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77
(3y−2)(3x+5)=77(3y−2)(3x+5)=77
⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77
Ta có bảng giá trị sau:
Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}
\(\left(3\cdot y\right)\cdot2-7=29\)
\(\Rightarrow6y=29+7\)
\(\Rightarrow6y=36\Rightarrow y=\frac{36}{6}=6\)
Vậy y = 6
2(3\(\times\)y)-7=29
\(\Leftrightarrow\)6y=36
\(\Leftrightarrow\)y=6
easy