Bài 1:

a) -16 +(x-3)²

<=> (x-3)²-16

<=> (x-3)² -4²

<=> (x-3-4)(x-3+4)

<=> (x-7)(x+1)

b) 64+16y+y²

<=> y² + 2.8.y + 8²

<=> (y+8)²

c) \(\dfrac{1}{8}-8x^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)

d)\(x^2-x+\dfrac{1}{4}\)

\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)

e) x⁴ + 4x² + 4

<=> (x²)² + 2.2.x² +2²

<=> (x² + 2)²

g)\(8x^3+60x^2y+150xy^2+125y^3\)

\(\Leftrightarrow\left(2x+5y\right)^3\)

Ban giup minh bai 2 luon voi nha Hậu Trần Công

1,Thực hiện phép tính :

a, (x + 2)⁹ : (x + 2)⁶

=(x+2)^9-6

=(x+2)³

b, (x - y) ⁴ : (x - 2)³

=(x-y)^4-3

=x-y

c, ( x²+ 2x + 4)⁵ : (x² + 2x + 4)

=(x²+2x+4)^5-1

=(x²+2x+4)⁴

d, 2(x² + 1)³ : 1/3(x² + 1)

=(2÷1/3).[(x²+1)³÷(x²+1)]

=6(x²+1)²

e, 5 (x - y)⁵ : 5/6 (x - y)²

=(5÷5/6).[(x-y)⁵÷(x-y)²]

=6(x-y)⁾³

Giúp mình nhé mọi người !

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

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

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

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

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

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

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

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

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

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

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

\(=\dfrac{2y}{\left(x-y\right)}\)

Tương tự các câu còn lại

Vậy (x^4 - x^3 - 3x^2 + x + 2) = (x^2 - x - 1)(x^2 - 1) + 1

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

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

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

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

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x+10=t\), ta có:

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

\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)

Giúp vs

a) \(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = 3 x y \left(\right. x + y \left.\right)\)

Giải:

Bắt đầu với vế trái của phương trình:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right)\)

Bước 1: Mở rộng \(\left(\right. x + y \left.\right)^{3}\):

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

Bước 2: Mở rộng \(\left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right)\):

\(\left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = x \left(\right. x^{2} - x y + y^{2} \left.\right) + y \left(\right. x^{2} - x y + y^{2} \left.\right)\)\(= x^{3} - x^{2} y + x y^{2} + y x^{2} - x y^{2} + y^{3}\)\(= x^{3} + y^{3} + \left(\right. y x^{2} - x^{2} y \left.\right) = x^{3} + y^{3}\)

Bước 3: Trừ các biểu thức:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = \left(\right. x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} \left.\right) - \left(\right. x^{3} + y^{3} \left.\right)\)\(= 3 x^{2} y + 3 x y^{2}\)\(= 3 x y \left(\right. x + y \left.\right)\)

Vậy, phương trình đã đúng:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = 3 x y \left(\right. x + y \left.\right)\)

b) \(B = \left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right) - 3 \left(\right. 9 x^{3} - 2 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right)\):

\(\left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right) = 3 x \left(\right. 9 x^{2} - 6 x + 4 \left.\right) + 2 \left(\right. 9 x^{2} - 6 x + 4 \left.\right)\)\(= 27 x^{3} - 18 x^{2} + 12 x + 18 x^{2} - 12 x + 8\)\(= 27 x^{3} + 8\)

Bước 2: Mở rộng \(3 \left(\right. 9 x^{3} - 2 \left.\right)\):

\(3 \left(\right. 9 x^{3} - 2 \left.\right) = 27 x^{3} - 6\)

Bước 3: Trừ hai biểu thức:

\(B = \left(\right. 27 x^{3} + 8 \left.\right) - \left(\right. 27 x^{3} - 6 \left.\right) = 8 + 6 = 14\)

Vậy, \(B = 14\).

c) \(C = \left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right) - \left(\right. x^{3} - 7 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right)\):

\(\left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right) = x \left(\right. x^{2} - 2 x + 4 \left.\right) - 2 \left(\right. x^{2} - 2 x + 4 \left.\right)\)\(= x^{3} - 2 x^{2} + 4 x - 2 x^{2} + 4 x - 8\)\(= x^{3} - 4 x^{2} + 8 x - 8\)

Bước 2: Trừ biểu thức \(x^{3} - 7\):

\(C = \left(\right. x^{3} - 4 x^{2} + 8 x - 8 \left.\right) - \left(\right. x^{3} - 7 \left.\right)\)\(C = x^{3} - 4 x^{2} + 8 x - 8 - x^{3} + 7\)\(C = - 4 x^{2} + 8 x - 1\)

Vậy, \(C = - 4 x^{2} + 8 x - 1\).

d) \(D = \left(\right. x + 1 \left.\right)^{3} - \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) - 3 x \left(\right. x + 1 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. x + 1 \left.\right)^{3}\):

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

Bước 2: Mở rộng \(\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\):

\(\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) = x \left(\right. x^{2} + x + 1 \left.\right) - 1 \left(\right. x^{2} + x + 1 \left.\right)\)\(= x^{3} + x^{2} + x - x^{2} - x - 1\)\(= x^{3} - 1\)

Bước 3: Mở rộng \(3 x \left(\right. x + 1 \left.\right)\):

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

Bước 4: Trừ các biểu thức:

\(D = \left(\right. x^{3} + 3 x^{2} + 3 x + 1 \left.\right) - \left(\right. x^{3} - 1 \left.\right) - \left(\right. 3 x^{2} + 3 x \left.\right)\)\(D = x^{3} + 3 x^{2} + 3 x + 1 - x^{3} + 1 - 3 x^{2} - 3 x\)\(D = 2\)

Vậy, \(D = 2\).

e) \(E = 3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) + x \left(\right. x + 1 \left.\right) - x \left(\right. x^{2} + x + 1 \left.\right)\)

Giải:

Bước 1: Mở rộng \(3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\):

\(3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) = 3 \left(\right. x \left(\right. x^{2} + x + 1 \left.\right) - \left(\right. x^{2} + x + 1 \left.\right) \left.\right)\)\(= 3 \left(\right. x^{3} + x^{2} + x - x^{2} - x - 1 \left.\right) = 3 \left(\right. x^{3} - 1 \left.\right)\)\(= 3 x^{3} - 3\)

Bước 2: Mở rộng \(x \left(\right. x + 1 \left.\right)\):

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

Bước 3: Mở rộng \(x \left(\right. x^{2} + x + 1 \left.\right)\):

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

Bước 4: Trừ các biểu thức:

\(E = \left(\right. 3 x^{3} - 3 \left.\right) + \left(\right. x^{2} + x \left.\right) - \left(\right. x^{3} + x^{2} + x \left.\right)\)\(E = 3 x^{3} - 3 + x^{2} + x - x^{3} - x^{2} - x\)\(E = 2 x^{3} - 3\)

Vậy, \(E = 2 x^{3} - 3\).

g) \(9 x \left(\right. x + 1 \left.\right)^{3} + \left(\right. x - 1 \left.\right)^{3} = 2 x^{3}\)

Giải:

Mở rộng biểu thức và kiểm tra tính đúng đắn:

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

Cộng cả hai biểu thức:

\(9 x \left(\right. x + 1 \left.\right)^{3} + \left(\right. x - 1 \left.\right)^{3} = 9 x^{4} + 27 x^{3} + 27 x^{2} + 9 x + x^{3} - 3 x^{2} + 3 x - 1\)\(= 9 x^{4} + 28 x^{3} + 24 x^{2} + 12 x - 1\)

So với \(2 x^{3}\), ta thấy biểu thức không đúng. Có thể bài toán có lỗi. Nếu có sự nhầm lẫn, bạn có thể điều chỉnh lại nhé!

h) \(\left(\right. x + 3 \left.\right) \left(\right. x^{2} - 3 x + 9 \left.\right) = x \left(\right. x^{2} - 3 x + 9 \left.\right) = x \left(\right. x^{2} + 4 \left.\right) - 1\)

a: \(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)

=-65

b \(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)

=27

c: \(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

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

\(=-3x^2+3x-3x+3x^2=0\)