câu a đề có sai số mũ ko vậy

b) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)

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

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

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

c) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

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

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

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

Lời giải:

a)

\(\frac{x-2}{6x^2-6x}-\frac{1}{4x^2-4}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x^2-1)}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x-1)(x+1)}\)

\(=\frac{2(x+1)(x-2)}{12x(x-1)(x+1)}-\frac{3x}{12x(x-1)(x+1)}=\frac{2(x+1)(x-2)-3x}{12x(x-1)(x+1)}\)

\(=\frac{2x^2-5x-4}{12x(x-1)(x+1)}=\frac{2x^2-5x-4}{12x^3-12x}\)

b) ĐK: \(x\neq \pm 1\)

\(\frac{(x+1)(x^2-2x+1)}{6x^3+6}:\frac{x^2-1}{4x^2-4x+4}\)

\(=\frac{(x+1)(x-1)^2}{6(x^3+1)}.\frac{4x^2-4x+4}{x^2-1}\)

\(=\frac{4(x+1)(x-1)^2(x^2-x+1)}{6(x+1)(x^2-x+1)(x^2-1)}\)

\(=\frac{2(x-1)}{3(x+1)}\)

Cảm ơn bn nha

1)

\(15x^3+29x^2-8x-12=(15x^3+30x^2)-(x^2+2x)-(6x+12)\)

\(=15x^2(x+2)-x(x+2)-6(x+2)\)

\(=(x+2)(15x^2-x-6)=(x+2)(15x^2-10x+9x-6)\)

\(=(x+2)[5x(3x-2)+3(3x-2)]\)

\(=(x+2)(3x-2)(5x+3)\)

2)

\(x^3+4x^2-29x+24=(x^3-x^2)+(5x^2-5x)-(24x-24)\)

\(=x^2(x-1)+5x(x-1)-24(x-1)\)

\(=(x-1)(x^2+5x-24)\)

\(=(x-1)(x^2-3x+8x-24)\)

\(=(x-1)[x(x-3)+8(x-3)]=(x-1)(x-3)(x+8)\)

a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)

Câub mô

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

\(a.17+8x=10-6x\\\Leftrightarrow 8x+6x=-17+10\\\Leftrightarrow 2x=-7\\ \Leftrightarrow x=-\frac{7}{2}\)

Vậy nghiệm của phương trình trên là \(-\frac{7}{2}\)

\(b.3\left(x+5\right)+7=19-5\left(x-2\right)\\\Leftrightarrow 3x+15+7=19-5x+10\\ \Leftrightarrow3x+5x=-15-7+19+10\\ \Leftrightarrow8x=7\\\Leftrightarrow x=\frac{7}{8}\)

Vậy nghiệm của phương trình trên là \(\frac{7}{8}\)

\(c.3x-4\left(x+2\right)\left(x+3\right)=14-4\left(x^2-3x\right)\\ \Leftrightarrow3x-4\left(x^2+5x+6\right)=14-4x^2+12x\\ \Leftrightarrow4x^2-4x^2+3x-5x-12x=24+14\\ \Leftrightarrow-14x=38\\ \Leftrightarrow x=-\frac{19}{7}\)

Vậy nghiệm của phương trình trên là \(-\frac{19}{7}\)

\(d.x+\frac{3}{4}+3x+2=\frac{x}{3}-3x-\frac{2}{6}\\ \Leftrightarrow\frac{12x}{12}+\frac{9}{12}+\frac{36x}{12}+\frac{24}{12}=\frac{4x}{12}-\frac{36x}{12}-\frac{4}{12}\\ \Leftrightarrow12x+9+36x+24=4x-36x-4\\ \Leftrightarrow12x+36x+36x-4x=-24-9-4\\ \Leftrightarrow80x=-37\\ \Leftrightarrow x=-\frac{37}{80}\)

a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0

=> x³ + 4x² - 29x + 24 = x²(x-1) + 5x(x-1) - 24(x-1)

= (x-1)(x²+5x-24) = (x-1)(x-3)(x+8)

b) ...

a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)

b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)

1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3) 
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8 
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5 
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27 
<=> - 3x^2 - 18x - 22 = 0 
<=> 3x^2 + 18x + 22 = 0 

Nửa chu vi mảnh đất là: 

                                               120 : 2 = 60 (m)

Chiều dài hơn chiều rộng là:

                                               5 + 5 = 10 (m)

Chiều rộng là:

                                          ( 60 - 10 ) : 2 = 25 (m)

Chiều dài là:

                                                25 + 10 = 35 (m)

Diện tích là:

                                               25  35 = 875 ( )