\(A=\left(2x\right)^2-2.2x.5+5^2-4x.x+4x.6\)

\(=4x^2-20x+25-4x^2+24x=4x+25\)

\(B=\left(7x-3y\right)^2-\left(7x-3y\right)\left(7x+3y\right)\)

\(=\left(7x-3y\right)\left(7x-3y-7x-3y\right)\)

\(=\left(7x-3y\right)\left(-6y\right)=18y^2-42xy\)

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

\(=9-2.3.2x+4x^2+9+2.3.2x+4x^2\)

\(=18+8x^2\)

\(D=\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+x\right)\left(y-z\right)\)

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

..........................

a)\(a^4+a^3+a^3b+a^2b=\left(a^4+a^3b\right)+\left(a^3+a^2b\right)\)

\(=a^3\left(a+b\right)+a^2\left(a+b\right)\)

\(=\left(a^3+a^2\right)\left(a+b\right)\)

\(=a^2\left(a+1\right)\left(a+b\right)\)

b)\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left[\left(x-y+4\right)-\left(2x+3y-1\right)\right]\left[\left(x-y+4\right)+\left(2x+3y-1\right)\right]\)

\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)

\(=\left(-x-4y+5\right)\left(4x+2y+3\right)\)

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

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

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

\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)

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

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

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

`@` `\text {Ans}`

`\downarrow`

\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)

`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`

`= x^2 - xy + xy - y^2 + y^2 - x^2`

`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`

`= 0`

Bài làm

Ta có: P = x³ + x²y - 2x² - xy - y² + 3y + x + 2017

          P = x³ + x²y - 2x² - xy - y² + 2y + y + x + 2017

          P = ( x³ + x²y − 2x² ) − ( xy + y² − 2y ) + ( x + y − 2 ) + 2019

          P = x²( x + y − 2 ) − y( x + y − 2 ) + ( x + y − 2 ) + 2019

Mà x + y = 2 => x + y - 2 = 0

Thay x + y - 2 = 0 và đa thức P, ta được:

P = x² . 0 - y . 0 + 0 + 2019

P = 0 - 0 + 0 + 2019

P = 2019

Vậy P = 2019 tại x + y = 2

# Học tốt #

\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(P=\left(x^3+x^2y-2x^2\right)+\left(-xy-y^2+2y\right)+\left(x+y-2\right)+2019\)

\(P=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(P=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)

\(P=0+2019=2019\)

Ta có

\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(\Leftrightarrow x^3+x^2y-2x^2-xy-y^2+2y+y+x+2017\)

\(\Leftrightarrow\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+2019\)

\(\Leftrightarrow x^2\cdot\left(x+y-2\right)-y\cdot\left(x+y-2\right)+\left(x+y-2\right)+2019\)

Ta có \(x+y=2\Rightarrow x+y-2=0\)

\(\Rightarrow P=2019\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)

\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)

\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)

a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)

\(=4x^2-4x+1-x^2+9-1969\)

\(=3x^2-4x-1959\)

b: \(\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\)

\(=4x^2-9y^2-4x^2+4xy-y^2\)

\(=-10y^2+4xy\)

a)\(\text{( 2 x − 1 )^2− ( x − 3 ) ( x + 3 ) − 1969}\)

\(\text{= 4x^2 − 4x + 1 − x^2 + 9 − 1969}\)

\(\text{=3x^2− 4 x − 1959}\)

b) \(\text{( 2 x − 3 y ) ( 2 x + 3 y ) − ( 2 x − y )^2}\)

=\(\text{= 4 x^2− 9 y^2− 4 x^2 + 4 x y − y^2}\)

\(\text{= -10 y^2+ 4 x y = -2 y ( 5 y -2 x )}\)

c)\(\text{( x + 3 y )^2 + ( x + y ) ( x − y ) + 280}\)

\(\text{= x^2 + 6 x y + 9 y^2 + x^2 − y^2 + 280}\)

\(\text{= 2 x^2 + 8 y^2 + 6 x y + 280}\)