Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

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

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

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

5x .(x\(^2\)- + \(\dfrac{1}{5}\)) = 5x\(^3\) - 15x\(^2\)+x

(x-3) (2x-1) =2x\(^2\) - x - 6x + 3 = 2x\(^2\) -7x +3

x\(^2\).(5x\(^3\)- x - \(\dfrac{1}{2}\) ) = 5x\(^5\)- x\(^3\)- \(\dfrac{1}{2}\)x\(^2\)

(x-2) (6x\(^2\) - 5x +1 )= 6x\(^3\)-5x\(^2\)+x -12x\(^2\) + 10x -2 =6x\(^3\) - 17x\(^2\) + 11x -2

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

14.C

15.D

Câu 14: C

Câu 15: D

14:A
15:D

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

a)\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

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

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

A. Cách B sai vì 5 : 2/5 thì ko thể nào = 25 đc.

a) Ta có: \(\dfrac{1-x}{x^2-2x+1}+\dfrac{x+1}{x-1}\)

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

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

\(=\dfrac{1-x-1}{1-x}=\dfrac{-x}{1-x}=\dfrac{x}{x-1}\)

b) Ta có: \(\dfrac{2x}{3y^4z}\cdot\left(-\dfrac{4y^2z}{5x}\right)\cdot\left(-\dfrac{15y^3}{8xz}\right)\)

\(=\dfrac{2x\cdot4y^2z\cdot15y^3}{3y^4z\cdot5x\cdot8xz}\)

\(=\dfrac{120xy^5z}{120x^2y^4z^2}=\dfrac{y}{xz}\)