b: =>(2y+1)(x-1)=8

mà x,y là số nguyên

nên \(\left(x-1\right)\left(2y+1\right)\in\left\{\left(8;1\right);\left(-8;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(9;0\right);\left(-7;-1\right)\right\}\)

c: \(\Leftrightarrow x\left(3y+1\right)-2\left(3y+1\right)=9\)

\(\Leftrightarrow\left(x-2\right)\left(3y+1\right)=9\)

\(\Leftrightarrow\left(x-2;3y+1\right)=\left(9;1\right)\)

hay (x,y)=(11;0)

D  nhá!!

(-3*x-1)*y+3*x^2+x

\(\left(3x+1\right)^2-\left(3x-1\right)^2\)

\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

\(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\cdot\left(x^2+3y^2\right)\)

______

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)

a)-(x-y)(x²+xy-1)=-(x³+x²y-x-x²y-xy²+y)

                          =-(x³-xy²-x+y)

                          =-x³+xy²+x-y

b)x²(x-1)-(x³+1)(x-y)=x³-x²-x³+x²y-x+y

                                =-x²+x²y-x+y

c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x²-3x-4x+2-15x²-10x-3x-2

                                             =-9x²-20x

d) hình như bạn ghi lỗi

Bài 2: C=x(x²-y)-x²(x+y)+y(x²-x)

             =x³-xy-x³-x²y+x²y-xy

             =-2xy

Thay x=1/2,y=-1 vào C, ta có:

        C=-2.1/2.(-1)=1

Vậy C=1 khi x=1/2 và y=-1.

a) Ta có: \(\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)

\(=\left(3x-2+3x+2\right)^2\)

\(=36x^2\)(1)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:

\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)

b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=100^2=10000\)

Chúc bạn học tốt 🙆‍♀️❤

Tiếp thu và sửa đổi. Cảm ơn ạ 🙆‍♀️

\(a) (x+y)+(y-x)\)

\(= x+y+y-x\)

\(=(x-x)+(y+y)\)

\(= 2y\)

\(b) (x+y)-(y-x)\)

\(= x+y-y+x\)

\(= (x+x)+(y-y)\)

\(= 2x\)

\(c) (x-y)+(y-x)\)

\(= x-y+y-x\)

\(= (x-x)+(y-y) \)

\(=0\)

\(e) (x^2+xy-1)+(3x^2+xy+1)\)

\(= x^2+xy-1+3x^2+xy+1\)

\(= (x^2+3x^2)+(xy+xy)+(1-1)\)

\(= 4x^2 + 2xy\)

\(g) (x^2+xy-1)-(3x^2+xy+1)\)

\(= x^2+xy-1-3x^2-xy-1\)

\(= (x^2-3x^2)+(xy-xy)-(1+1)\)

\(= -2x^2 - 2\)

\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)

\(x^3-3x^2y+x+3xy^2-y-y^3\\=(x^3-3x^2y+3xy^2-y^3)+(x-y)\\=(x-y)^3+(x-y)\\=(x-y)[(x-y)^2+1]\\=(x-y)(x^2-2xy+y^2+1)\)