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a) A + x2 - 4xy2 + 2xz - 3y2 = 0
=> A = -x2 + 4xy2 - 2xz + 3y2
b) B + 5x2 - 2xy = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - 5x2 + 2xy= x2 + 11xy - y2
c) 3xy - 4y2 - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - x2 + 7xy - 8y2 = -12y2 + 10xy - x2
Trả lời:
a, A + ( x2 - 4xy2 + 2xz - 3y2 ) = 0
=> A = - ( x2 - 4xy2 + 2xz - 3y2 ) = - x2 + 4xy2 - 2xz + 3y2
b, B + ( 5x2 - 2xy ) = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - ( 5x2 - 2xy ) = 6x2 + 9xy - y2 - 5x2 + 2xy = x2 + 11xy - y2
c, ( 3xy - 4y2 ) - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - ( x2 - 7xy + 8y2 ) = 3xy - 4y2 - x2 + 7xy - 8y2 = 10xy - 12y2 - x2
d, B + ( 4x2y + 5y2 - 3xz + z2 ) = x2 + 11xy - y2 + 4x2y + 5y2 - 3xz + z2 = x2 + 11xy + 4y2 + 4x2y - 3xz + z2
Q = x2 + y2 + z2 + x2 – y2 + z2 + x2 + y2 - z2
= x2 + x2 + x2 + y2 + y2 - y2 + y2 + z2 + z2 - z2
= 3x2 + y2 + z2
\(x^2+y^2+z^2+x^2-y^2+z^2+x^2+y^2-z^2\)
\(=x^2+x^2+x^2+y^2+y^2-y^2+z^2+z^2-z^2\)
\(=3x^2+y^2+z^2\)
A + B = (2x^2 y^2 - 4x^3 + 7xy - 18) + (x^3y + x^2y^2 - 15xy + 1)
= 2x^2 y^2 - 4x^3 + 7xy - 18 + x^3y + x^2y^2- 15xy + 1
= (2x^2 y2 + x^2y^2) - 4x^3 + x^3y + (7xy – 15xy) + ( -18 + 1)
= 3x^2 y2 - 4x^3 + x^3y – 8xy – 17
c) \(x+y+9=xy-7\)
=> \(x+y+16=xy=>x+16=xy-y=y.\left(x-1\right)\)
\(=>y=\frac{x+16}{x-1}\) (x khác 1)
Mà do y thuộc Z => \(\frac{x+16}{x-1}\in Z=>x+16⋮x-1=>\left(x-1\right)+17⋮x-1=>x-1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
\(=>x\in\left\{0;2;-16;18\right\}\) (Thỏa mãn do khác 1)
*) Nếu x=0 => 16+y=0=> y=-16.
*) Nếu x=2 => 18+y=2y=> y=18
*) Nếu x=-16 => y=-16y => y=0
*) Nếu x=18 => y=2
Vậy (x,y)=.....
a) Ta có M + (5x2 - 2xy) = 6x2 + 9xy - y2
=> M = 6x2 + 9xy - y2 - (5x2 - 2xy) = x2 + 11xy - y2
b) Ta có M - (3xy - 4y2) = x2 - 7xy + 8y2
=> M = 3xy - 4y2 + x2 - 7xy + 8y2 = 4y2 - 4xy + x2
c) Ta có (25x2y - 13xy + y3) - M = 11x2y - 2y2
=> M = (25x2y - 13xy + y3) - (11x2y - 2y2) = 14x2y - 13xy + y3 + 2y2
d) Ta có M + (12x4 - 15x2y + 2xy2 + 7) = 0
=> M = -12x4 + 15x2y - 2xy2 - 7
\(B=3+3^3+3^5+...+3^{101}\)
\(3^2.B=3^3+3^5+3^7+...+3^{103}\)
\(\left(3^2-1\right)B=\left(3^3+3^5+3^7+...+3^{103}\right)-\left(3+3^3+3^5+...+3^{101}\right)\)
\(8B=3^{103}-3\)
\(B=\frac{3^{103}-3}{8}\)
a: Sửa đề: \(25-y^2=8\left(x-2009\right)^2\)
Ta có: \(25-y^2=8\left(x-2009\right)^2\)
=>\(25-y^2\) ⋮8 và \(25-y^2\ge0\)
=>\(y^2\in\left\lbrace1;9;25\right\rbrace\)
TH1: \(y^2=1\)
\(25-y^2=8\left(x-2009\right)^2\)
=>\(8\left(x-2009\right)^2=25-1=24\)
=>\(\left(x-2009\right)^2=3\)
mà x nguyên
nên x∈∅
=>Loại
TH2: \(y^2=9\)
\(25-y^2=8\left(x-2009\right)^2\)
=>\(8\left(x-2009\right)^2=25-9=16\)
=>\(\left(x-2009\right)^2=2\)
mà x nguyên
nên x∈∅
=>Loại
TH3: \(y^2=25\)
\(25-y^2=8\left(x-2009\right)^2\)
=>\(8\left(x-2009\right)^2=25-25=0\)
=>\(\left(x-2009\right)^2=0\)
=>x-2009=0
=>x=2009(nhận)
\(y^2=25\)
=>y=5(nhận) hoặc y=-5(nhận)
b: \(x^3y=xy^3+1997\)
=>\(x^3y-xy^3=1997\)
=>\(xy\left(x^2-y^2\right)=1997\)
=>xy(x-y)(x+y)=1997
Đặt A=xy(x-y)(x+y)
TH1: x lẻ; y lẻ
=>x-y chẵn
=>xy(x-y)(x+y)⋮2
=>A⋮2(1)
TH2: x lẻ; y chẵn
=>xy⋮2
=>xy(x-y)(x+y)⋮2
=>A⋮2(2)
TH3: x chẵn; y lẻ
=>xy⋮2
=>xy(x-y)(x+y)⋮2
=>A⋮2(3)
TH4: x chẵn; y chẵn
=>xy⋮2
=>xy(x-y)(x+y)⋮2
=>A⋮2(4)
Từ (1),(2),(3),(4) suy ra A⋮2
mà 1997 lẻ
nên (x;y)∈∅
c: x+y+9=xy-7
=>xy-7-x-y-9=0
=>xy-x-y-16=0
=>x(y-1)-y+1-17=0
=>(x-1)(y-1)=17
=>(x-1;y-1)∈{(1;17);(17;1);(-1;-17);(-17;-1)}
=>(x;y)∈{(2;18);(18;2);(0;-16);(-16;0)}
may hoi the chiu
an ơi m đao ak