a/ Vế trái = x³ + x² + x - x² - x - 1 = x³ - 1 (= vế phải)

b/ hình như đề sai

b/ Vế trái = x⁴ + x³y + x²y² + xy³ - x³y - x²y² - xy³ - y⁴ = x⁴ - y⁴ (= vế phải)

b)(x^3+x^2y+xy^2+y^3)(x-y)=x^4-y^4

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

\(=\left(2x+y\right).3y\)

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

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)

\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

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

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

\(=x\left(x-y\right)^2-4x\)

\(=x\left[\left(x-y\right)^2-4\right]\)

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

a) (x + 2y)² - (x - y)²

= (x + 2y - x + y)(x + 2y + x - y)

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

= x(x - y - 2)(x - y + 2)

a)\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)

\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)

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

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

b)\(\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)

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

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

a) (x - y)(x⁴ + x³y + x²y² + xy³ + y⁴)

= x(x⁴ + x³y + x²y² + xy³ + y⁴) - y(x⁴ + x³y + x²y² + xy³ + y⁴)

= x⁵ + x⁴y + x³y² + x²y³ + xy⁴ - x⁴y - x³y² - x²y³ - xy⁴ - y⁵

= x⁵ - y⁵

b) (a + b)(a² - ab + b²)

= a(a² - ab + b²) + b(a² - ab + b²)

= a³ - a²b + ab² + a²b - ab² + b³

= a³ + b³

a: C=-2x^4+3x^2y-2xy+y^2+7

Bậc là 4

b: B=5x^4-3x^2y+2xy+y^2

D=-2x^4+3x^2y-2xy+y^2+7+5x^4-3x^2y+2xy+y^2

=3x^4+2y^2

E=-2x^4+3x^2y-2xy+y^2+7-5x^4+3x^2y-2xy-y^2

=-7x^4+6x^2y-4xy+7

\(a.2x^3+6x=2x\left(x^2+3\right)\)

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

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

\(b.5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)\)

\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)

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

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

\(d.y^2\left(x^2+y\right)-x^3-xy\)

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

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

e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu

\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)

\(=3x^2\left(y^2-2x\right)+15x\left(y^2-2x\right)\)

\(=\left(y^2-2x\right)\left(3x^2+15x\right)\)

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

a) 2x=3y=4z⇒\(\dfrac{2}{\dfrac{1}{x}}=\dfrac{3}{\dfrac{1}{y}}=\dfrac{4}{\dfrac{1}{z}}=\dfrac{2+3+4}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{9}{3}=3\) ( Vì\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\))

⇒ x=\(\dfrac{3}{2}\) ; y=1; z=\(\dfrac{3}{4}\)

b) \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-by}{c}\)

= \(\dfrac{abz-acy}{a^{2^{ }}}=\dfrac{bcx-abz}{b^{2^{ }}}=\dfrac{acy-bcy}{c^2}\) =\(\dfrac{\left(abz-acy\right)+\left(bcx-abz\right)+\left(acy-bcy\right)}{a^2+b^2+c^{2^{ }}}=\dfrac{0}{a^2+b^2+c^{2^{ }}}=0\)

⇒\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-by}{c}=0\)

⇒ ✽bz-cy=0⇒bz=cy⇒\(\dfrac{b}{y}=\dfrac{c}{z}\) (1)

✽ cx-az=0⇒cx=az⇒ \(\dfrac{a}{x}=\dfrac{c}{z}\) (2)

Từ (1) và (2) suy ra\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)

d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)

e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)

f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)

g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)