Bài 1 : Phân tích các đa thức sau thành nhân tử :

a) 8x³ - 64

=(2x)³ + 4³

=(2x+4)(4x² - 8x + 16)

c) 125x³ + 1

=5x³ + 1³

=(5x+1)(25x² +5x+1)

d) 8x³ - 27

=(2x)³ - 3³

=(2x - 3)(2x² + 6x + 9)

e) 1 + 8x⁶y³

=1 + (2x²y)³

=(1 + 2x²y)(4x⁴y² -2x²y + 1)

f) 125x³ + 27y³

=(5x)³ + (3y³)

=(5x + 3y)(25x² - 15xy + 9y²)

Bài 1

a) \(8x^3-64\)

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

\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)

c) \(125x^3+1\)

\(=\left(5x\right)^3+1^3\)

\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)

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

\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)

e) \(1+8x^6x^3\)

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

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

f) \(125x^3+27y^3\)

\(=\left(5x\right)^3+\left(3y\right)^3\)

\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)

Bài 2

b) \(x^3+6x^2+12x+8\)

\(=x^3+3.x^2.2+3.x.2^2+2^3\)

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

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

\(=x^3-3.x^2.1+3.x.1^2-1^3\)

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

d) \(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3.\left(3x\right)^2.2y+3.3x.\left(2y\right)^2-\left(2y\right)^3\)

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

a) x⁴ - y⁴

= (x²)² - (y²)²

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

a.\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)g,\(27x^3-0.001=27x^3-0=27x^3\)

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

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

e: \(=\left(2x-1\right)^2-\left(x+1\right)^2\)

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

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

g: \(=\left(3x-\dfrac{1}{10}\right)\left(9x^2+\dfrac{3}{10}x+\dfrac{1}{100}\right)\)

k: \(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

Câu 2: 

a) Để đồ thị hàm số \(y=\left(m+1\right)x^2\) đi qua điểm A(1;2) thì

Thay x=1 và y=2 vào hàm số \(y=\left(m+1\right)x^2\), ta được:

m+1=2

hay m=1

Vậy: m=1

a) Ta có: \(x^2+2x+1\)

\(=x^2+2\cdot x\cdot1+1^2\)

\(=\left(x+1\right)^2\)

b) Ta có: \(1-2y+y^2\)

\(=y^2-2\cdot y\cdot1+1^2\)

\(=\left(y-1\right)^2\)

c) Ta có: \(x^3-3x^2+3x-1\)

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

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

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

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

d) Ta có: \(27+27x+9x^2+x^3\)

\(=x^3+3x^2+6x^2+18x+9x+27\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+9\right)\)

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

e) Ta có: \(8-125x^3\)

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

\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)

f) Ta có: \(64x^3+\frac{1}{8}\)

\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)

\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)

g) Ta có: \(1-x^2y^4\)

\(=1^2-\left(xy^2\right)^2\)

\(=\left(1-xy^2\right)\left(1+xy^2\right)\)

a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)

b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)

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

d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)

e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)

f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)

Ko chắc ạ!

Đề bài : Tính?

125.(-61).(-2).(-1)²ⁿ  (n thuộc N*)

=125.(-61).(-2).1

=[125.(-2)].(-61)

=(-250).(-61)=15250

\(125.\left(-61\right).\left(-2\right).\left(-1\right)^{2n}\)

\(=125.\left(-61\right).\left(-2\right).1\)(vì số mũ của -1 là số chẵn)

\(=125.122\)

\(=15250\)

Bài I :

1 ) \(3x\left(x-5\right)-\left(3x+2\right)\left(3x-2\right)=31\)

\(\Leftrightarrow3x^2-15x-9x^2+4-31=0\)

\(\Leftrightarrow-6x^2-15x-27=0\)

Phương trình vô nghiệm .

2 )

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=16\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

Bài II :

\(B=n\left(n+5\right)-\left(n-3\right)\left(n+20\right)\)

\(=n^2+5n-n^2-17n+60\)

\(=-12n+60\)

\(=-12\left(n-5\right)\)

Vì \(-12\) chia hết cho 6 \(\Rightarrow-12\left(n-5\right)\) chia hết cho 6 .

Vậy \(n\left(n+5\right)-\left(n-3\right)\left(n+20\right)\) chia hết cho 6 (đpcm)

a) 6x⁴ - 9x³ = 3x³(2x - 3)

b) 5y¹⁰ + 15y⁶ = 5y⁶(y⁴ + 3)

c) x² - 6xy + 9y² = x² - 6xy + (3y)² = (x - 3y)²

e) x³ - 64 = x³ - 4³ = (x - 4)(x² + 4x + 16)

f) 125x³ + y⁶ = (5x)³ + (y²)³ = (5x + y²)(25x² + 5y² + y⁴)

g) 0,125(a + 1)³ - 1 = [0,5(a + 1)]³ - 1³ = (0,5a + 0,5)³ - 1³ = (0,5a + 0,5 - 1)[(0,5a + 0,5)² + (0,5a + 0,5) + 1) = (0,5a - 0,5)(0,25a^2 + 0,5 a + 0,25 + 0,5a + 0,5 + 1) = (0,5a - 0,5)(0,25a² + 1,75 + a)

h) 3x² - 12y² = 3(x² - 4y²) = 3[x² - (2y)² ] = 3(x - 2y)(x + 2y)