a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

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

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)

\(=x^3+64\\ =x^3-27y^3\\ =x^6-\dfrac{1}{27}\)

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

\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)

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

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

a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

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

=0

b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)

\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)

\(=\dfrac{2}{27}\)

c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

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

=0

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x_1-1}{5}=\dfrac{x_2-2}{4}=\dfrac{x_3-3}{3}=\dfrac{x_4-4}{2}=\dfrac{x_5-5}{1}\)

\(=\dfrac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)

\(=\dfrac{\left(x_1+x_2+x_3+x_4+x_5\right)-\left(1+2+3+4+5\right)}{15}\)

\(=\dfrac{30-15}{15}=1\)

\(\Rightarrow x_1=x_2=x_3=x_4=x_5=6\)

Vậy...

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x1-1}{5}\)=\(\dfrac{x2-2}{4}\)\(\dfrac{x3-3}{3}\)=\(\dfrac{x4-4}{2}\)=\(\dfrac{x5-5}{1}\)=\(\dfrac{x1-1+x2-2+x3-3+x4-4+x5-5}{5+4+3+2+1}\)=\(\dfrac{x1+x2+x3+x4+x5-\left(1+2+3+4+5\right)}{15}\)=\(\dfrac{30-15}{15}\)=\(\dfrac{15}{15}\)=1

\(\dfrac{x1-1}{5}\)=1 => x1-1=5 => x1 =6

\(\dfrac{x2-2}{4}\)=1 => x2-2=4 => x2 =6

\(\dfrac{x3-3}{3}\)=1 => x3-3=3 => x3 =6

\(\dfrac{x4-4}{2}\)=1 => x4-4=2 => x4 =6

\(\dfrac{x5-5}{1}\)=1 => x5-5=1 => x5 = 6

Vậy x1=x2=x3=x4=x5 =6

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

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

\(\Leftrightarrow A=5x^2+y^2\)

\(B+\left(-x^4+x^2-2x^3-\dfrac{1}{3}\right)=3x^2-2x^3+x-\dfrac{2}{3}\)

\(\Leftrightarrow B=3x^2-2x^3+x-\dfrac{2}{3}+x^4-x^2+2x^3+\dfrac{1}{3}\)

\(\Leftrightarrow B=x^4+2x^2+x-\dfrac{1}{3}\)

A=5x²+y².

B=x⁴+2x²+x-1/3.

a)
\(=x^3+3.x^2.1+3.x.1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
c)
\(x^3+3.x^2.\dfrac{1}{2}+3.x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=x^3+1,5x^2+0,75x+0,125\)

d)
=\(\left(x^2\right)^3-3.\left(x^2\right)^2.2+3.x^2.2^2-2^3\)
\(=x^5-6x^4+12x^2-8\)
e)
\(=\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)

b) Đặt t = x² ( t ≥ 0) ta có pt:

t² - t² - 2= 0

Δ= (-1)² - 4.1. (-2)

  = 9 > 0

⇒ \(\sqrt{\Delta}=\sqrt{9}=3\)

Vậy pt có 2 n_o phân biệt

x₁= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-1\right)+3}{2.1}=2\)

x₂= \(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-1\right)-3}{2.1}=-1\)

Với t = 2 thì x²= 2 ⇔ x_1;2 = \(\pm4\)

Với t = -1 thì x²= -1 ⇔ x_3;4 ∈ ∅

Vậy tập nghiệm của pt là: S= \(\left\{\pm4\right\}\)

c) Đặt t = x² ( t ≥ 0) ta có pt:

4t² - 5t² - 9= 0

Δ= (-5)² - 4.4. (-9)

  = 169 > 0

⇒ \(\sqrt{\Delta}\) = \(\sqrt{169}=13\)

Vậy pt có 2 n_o phân biệt

x₁= \(\dfrac{5+13}{2.4}=\dfrac{9}{4}\)

x₂= \(\dfrac{5-13}{2.4}=-1\)

Với t = \(\dfrac{9}{4}\)  thì x²= \(\dfrac{9}{4}\) ⇔ x_1;2 = \(\pm\dfrac{3}{2}\)

Với t = -1 thì x²= -1 ⇔ x_3;4 ∈ ∅

Vậy tập nghiệm của pt là: S= \(\left\{\pm\dfrac{3}{2}\right\}\)

a: =>\(\dfrac{x+1-2x}{x\left(x+1\right)}=1\)

=>-x+1=x^2+x

=>x^2+x+x-1=0

=>x^2+2x-1=0

=>\(x=-1\pm\sqrt{2}\)

b: =>x^4+2x^2-x^2-2=0

=>(x^2+2)(x^2-1)=0

=>x^2-1=0

=>x^2=1

=>x=1 hoặc x=-1

c: =>4x^4-9x^2+4x^2-9=0

=>(4x^2-9)(x^2+1)=0

=>4x^2-9=0

=>x=3/2 hoặc x=-3/2

Sửa đề: \(P=3x^3+x^2+4x^4-x-3x^3+5x^4+x^2-6\)

Ta có: \(P=3x^3+x^2+4x^4-x-3x^3+5x^4+x^2-6\)

\(=9x^4+2x^2-x-6\)

Ta có: \(Q\left(x\right)=2x^3-x^4-\dfrac{1}{2}x^2-3+\dfrac{3}{4}x-\dfrac{1}{3}x^2+x^4-\dfrac{7}{4}x\)

\(=2x^3-\dfrac{5}{6}x^2-x-3\)