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a/ \(=x^3-2x^5\)

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

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)

2:

a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)

b: \(2\left(x-1\right)+x^2-x\)

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

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

c: \(3x^2+14x-5\)

\(=3x^2+15x-x-5\)

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

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

=>\(2x^2-2x-2x^2=4\)

=>-2x=4

=>x=-2

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

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

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

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)

=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)

=>4x(2x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)

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

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

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

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

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

=-6

\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)

\(=6x^2+12x+x+2-6x^2+10x\)

\(=23x+2\)

a) (6x + 1)(x + 2) - 2x(3x - 5)

= 6x² + 12x + x + 2 - 6x² + 10x

= (6x² - 6x²) + (12x + x + 10x) + 2

= 23x + 2

b) (2x - 1)² - (2x - 3)(2x + 3)

= 4x² - 4x + 1 - 4x² + 9

= (4x² - 4x²) - 4x + (1 + 9)

= -4x + 10

c) (2x - 3)³  - (3x  + 1)(5 - 4x) - 16x²

= 8x³ - 36x² + 54x - 15x + 12x² - 5 + 4x - 16x²

= 8x³ - (36x² - 12x² + 16x²) + (54x - 15x + 4x) - 5

= 8x³ - 40x² + 43x - 5

d) (3x + 2) - (x - 5) - x(3x - 13)

= 3x  + 2 - x + 5 - 3x² + 13x

= (3x - x + 13x) + (2 + 5) - 3x²

= 15x + 7 - 3x²

\(e,\left(5x+1\right)\left(x-2\right)-x\left(5x-9\right)\)

\(=5x^2-10x+x-2-5x^2+9x\)

\(=-2\)

\(A=\left(x+2\right)^2-\left(x+3\right)\left(x-1\right)+15\)

\(A=x^2+4x+4-\left(x^2-x+3x-3\right)+15\)

\(A=\left(x^2-x^2\right)+\left(4x+x-3x\right)+\left(15+3+4\right)\)

\(A=2x+22\)

______________________

\(B=\left(x+1\right)\left(x-1\right)-\left(x+4\right)^2-6\)

\(B=\left(x^2-1\right)-\left(x^2+8x+16\right)-6\)

\(B=\left(x^2-x^2\right)-8x-\left(1+16+6\right)\)

\(B=-8x-23\)

_________________

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

\(C=\left[\left(3x\right)^2-2^2\right]-\left(9x^2-6x+1\right)\)

\(C=\left(9x^2-9x^2\right)+6x-\left(4+1\right)\)

\(C=6x-5\)

a) Rút gọn biểu thức A = (x + 2)2 - (x + 3)(x - 1) + 15:

Bắt đầu bằng việc mở ngoặc:
A = (x^2 + 4x + 4) - (x^2 + 2x - 3x - 3) + 15

Tiếp theo, kết hợp các thành phần tương tự:
A = x^2 + 4x + 4 - x^2 - 2x + 3x + 3 + 15

Tiếp tục đơn giản hóa:
A = x^2 - x^2 + 4x - 2x + 3x + 4 + 3 + 15

Kết quả cuối cùng:
A = 5x + 19

b) Rút gọn biểu thức B = (x - 1)(x + 1) - (x + 4)2 - 6:

Bắt đầu bằng việc mở ngoặc:
B = (x^2 - 1) - (x^2 + 4x + 4) - 6

Tiếp theo, kết hợp các thành phần tương tự:
B = x^2 - 1 - x^2 - 4x - 4 - 6

Tiếp tục đơn giản hóa:
B = x^2 - x^2 - 4x - 4 - 6 - 1

Kết quả cuối cùng:
B = -4x - 11

c) Rút gọn biểu thức C = (3x - 2)(3x + 2) - (3x - 1)2:

Bắt đầu bằng việc mở ngoặc:
C = (9x^2 - 4) - (9x^2 - 6x + 1)

Tiếp theo, kết hợp các thành phần tương tự:
C = 9x^2 - 4 - 9x^2 + 6x - 1

Tiếp tục đơn giản hóa:
C = 9x^2 - 9x^2 + 6x - 4 - 1

Kết quả cuối cùng:
C = 6x - 5

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

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

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

\(=4x^2\)

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

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

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

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

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

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

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

\(=125-150x+60x^2-8x^3-\left(25-9x^2\right)\)

\(=125-150x+60x^2-8x^3-25+9x^2\)

\(=100-15x+69x^2-8x^3\)

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

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

\(=\left(x+2\right).5x\)

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

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

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

\(E=\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-8+x\left(x^2-1\right)+6x^2-18x\)

\(=x^3-6x^2+12x-8+x^3-x+6x^2-18x\)

\(=-7x-8\)

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

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

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

\(=-3\)