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1: xy+x+y+1=0
=>x(y+1)+(y+1)=0
=>(x+1)(y+1)=0
=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)
2: xy+x+6=0
=>x(y+1)=-6
=>(x;y+1)∈{(1;-6);(-6;1);(-1;6);(6;-1);(2;-3);(-3;2);(-2;3);(3;-2)}
=>(x;y)∈{(1;-7);(-6;0);(-1;5);(6;-2);(2;-4);(-3;1);(-2;2);(3;-3)}
3: -xy-x-y-1=0
=>xy+x+y+1=0
=>x(y+1)+(y+1)=0
=>(x+1)(y+1)=0
=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)
4: xy-x-y+1=0
=>x(y-1)-(y-1)=0
=>(x-1)(y-1)=0
=>\(\begin{cases}x-1=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=1\end{cases}\)
5: xy+2x+y+11=0
=>x(y+2)+y+2+9=0
=>x(y+2)+(y+2)=-9
=>(x+1)(y+2)=-9
=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}
=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}
6: ĐKXĐ: x<>0
\(\frac{5}{x}+\frac{y}{4}=\frac18\)
=>\(\frac{20+xy}{4x}=\frac18\)
=>\(\frac{40+2xy}{8x}=\frac{x}{8x}\)
=>40+2xy=x
=>x-2xy=40
=>x(1-2y)=40
=>x(2y-1)=-40
mà 2y-1 lẻ(do y nguyên)
nên (x;2y-1)∈{(-40;1);(40;-1);(8;-5);(-8;5)}
=>(x;2y)∈{(-40;2);(40;0);(8;-4);(-8;6)}
=>(x;y)∈{(-40;1);(40;0);(8;-2);(-8;3)}
8: (x+2)(y-3)=-3
=>(x+2;y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}
=>(x;y)∈{(-1;0);(-5;4);(-3;6);(1;2)}
Bài 1:
\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(\frac{2}{5}.\frac{-5}{4}\right)\)
\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)
\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y\right).\left(\frac{-3}{4}.\frac{-8}{9}\right)\)
\(B=x^8.y^7.\frac{2}{3}\)
Bài 2:
\(A=\left(15.x^2.y^3-12.x^2.y^3\right)+\left(11x^3.y^2-8.x^3.y^2\right)+\left(7x^2-12x^2\right)\)
\(A=3.x^2.y^3+2.x^3.y^2-5x^2\)
B tương tự nhé, đáp án là (theo mình)
\(B=\frac{5}{2}.x^5.y+\frac{7}{3}.x.y^4-\frac{1}{4}.x^2.y^3\)
1.
\((\frac{1}{3}xy)^2.x^3+\frac{3}{2}(2x)^3(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=(\frac{1}{9}x^2y^2)x^3+\frac{3}{2}(8x^3)(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}(x^2.x^3)y^2+(\frac{3}{2}.8.\frac{-7}{4})(x^3.x^2).y^2-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}x^5y^2-21x^5y^2-\frac{2}{3}x^5y^2=\frac{-194}{9}x^5y^2\)
2.
\(\frac{-2}{5}x^2y(-y^6)+\frac{3}{2}xy(\frac{-1}{15}xy^6)+(-2xy)^2y^5\)
\(=\frac{2}{5}x^2(y.y^6)+(\frac{3}{2}.\frac{-1}{15})(x.x).(y.y^6)+4x^2(y^2.y^5)\)
\(=\frac{2}{5}x^2y^7-\frac{1}{10}x^2y^7+4x^2y^7=\frac{43}{10}x^2y^7\)
3.
\(\frac{3}{7}xy^2z+\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2-\frac{3}{7}xy^2z\)
\(=(\frac{3}{7}xy^2z-\frac{3}{7}xy^2z)+(\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2)\)
\(=\frac{5}{6}x^3y^2\)
4.
\(\frac{2}{3}xy^2-\frac{5}{2}yz+\frac{1}{2}xy^2-\frac{2}{3}yz\)
\(=(\frac{2}{3}xy^2+\frac{1}{2}xy^2)-(\frac{5}{2}yz+\frac{2}{3}yz)\)
\(=\frac{7}{6}xy^2+\frac{19}{6}yz\)
5.
\(\frac{3}{2}xy^2z^5-\frac{5}{4}xyz^2+\frac{4}{3}xy^2z^5+\frac{1}{2}xyz^2\)
\(=(\frac{3}{2}xy^2z^5+\frac{4}{3}xy^2z^5)+(\frac{-5}{4}xyz^2+\frac{1}{2}xyz^2)\)
\(=\frac{17}{6}xy^2z^5-\frac{3}{4}xyz^2\)
Bài 2b
Thay x = -1; y = 1 vào N ta đc:
\(N=\left(-1\right).1+\left(-1\right)^2.1^2+\left(-1\right)^3.1^3+\left(-1\right)^4.1^4+\left(-1\right)^5.1^5\)
\(=\left(-1\right)+1+\left(-1\right)+1+\left(-1\right)\)
\(=-1\)
=(\(4x^2-x^2+4x^2\) ) + (\(-3xy-xy\) )+ (\(y^2+2y^2+3y^2\) )
=(\(7x^2\))+(\(-4xy\))+(\(6y^2\))
a, \(A=x^3-x^2y+3x^2-xy+y^2-4y+x+2\)
\(=x^3-x^2y+3x^2-\left(xy-y^2+3y\right)-y+x+3-1\)
\(=x^2\left(x-y+3\right)-y\left(x-y+3\right)+\left(x-y+3\right)-1\)
Thay x-y+3=0 vào A
\(A=x^2.0-y.0+0-1=-1\)
b, \(B=x^3-2x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y-x^2y+3x^2+xy^2-3xy-2y+2x+4\)
\(=x^3-x^2y+3x^2-x^2y+xy^2-3xy+2x-2y+6-2\)
\(=x^2\left(x-y+3\right)-xy\left(x-y+3\right)+2\left(x-y+3\right)-2\)
Thay x-y+3=0 vào B
\(B=x^2.0-xy.0+2.0-2=-2\)
a) A = x2 + 2xy – 3x3 + 2y3 + 3x3 – y3 tại x = 5 và y = 4.
Trước hết ta thu gọn đa thức
A = x2 + 2xy – 3x3 + 2y3 + 3x3 – y3 = x2 + 2xy + y3
Thay x = 5; y = 4 ta được:
A = 52 + 2.5.4 + 43 = 25 + 40 + 64 = 129.
Vậy A = 129 tại x = 5 và y = 4.
b) M = xy - x2y2 + x4y4 – x6y6 + x8y8 tại x = -1 và y = -1.
Thay x = -1; y = -1 vào biểu thức ta được:
M = (-1)(-1) - (-1)2.(-1)2 + (-1)4. (-1)4-(-1)6.(-1)6 + (-1)8.(-1)8
= 1 -1 + 1 - 1+ 1 = 1.
\(a.\)\(x^2+2xy-3x^3+2y^3+3x^3-y^3\)
=\(x^2+2xy+y^3\)
\(thếx=5;y=4\) \(ta\) \(có\)
= \(5^2+2.5.4+4^3\)
= 25 + 40 + 64
=129
b.
\(xy-x^2y^2+x^4y^4-x^6y^6+x^8y^8\)
thế \(x=-1;y=-1\) ta có:
(-1).(-1) - \(\left(-1\right)^2.\left(-1\right)^2\)+\(\left(-1\right)^4.\left(-1\right)^4-\left(-1\right)^6.\left(-1\right)^6+\left(-1\right)^8.\left(-1\right)^8\)
= 1 - 1.1 +1.1 - 1.1 +1.1
= 1-1+1-1+1
= 1
x6 + x2y5 + xy6 + x2y5 – xy6
= x6 + (x2y5 + x2y5) + (xy6 – xy6)
= x6 + (1 + 1)x2y5 + (1-1)xy6
= x6 + 2x2y5