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Câu 3: 

a: \(49^2=2401\)

b: \(51^2=2601\)

c: \(99\cdot100=9900\)

a) (x+2)²+x(x-4)

   =x²+4x+4+x²-4x

   =2x²+4

b)(x-3)²-(x+3)(x-4)

  =x²-6x+9-x²+4x-3x+12

  =-5x+12

c) (3x+1)²+3x(2-4x)

   =9x²+6x+1+6x-12x²

   =-3x²+12x+1

d) (2x-4y)²-(2x-3)(2x-3y)

  =4x²-16xy+16y²-4x²+6xy+6x-9y

  =16y²-10xy+6x-9y

a) (2x+3)²-2(2x+3)(2x+5)+(2x+5)²

=4x²+12x+9-(4x+6)(2x+5)+4x²+20x+25

=4x²+12x+9-(8x²+12x+20x+30)+4x²+20x+25

=4x²+12x+9-8x²-12x-20x-30+4x²+20x+25

=4

b) (x²+x+1)(x²-x+1)(x²-1)

=((x²+1)²-x²)(x²-1)

=(x⁴+x²+1)(x²-1)

=x⁶+x⁴+x²-x⁴-x²-1

=x⁶-1

c)(a+b-c)²+(a-b+c)²-2(b-c)²

=a²+b²+c²+2ab-2ac-2bc+a²+b²+c²-2ab+2ac-2bc-2(b²-2bc+c²)

=2a²+2b²+2c²-4bc-2b²+4bc-2c²

=2a²

d) (a+b+c)²+(a-b-c)²+(b-c-a)²+(c-a-b)²

= a²+b²+c²+2ab+2ac+2bc+a²+b²+c²-2ab-2ac+2bc+a²+b²+c²+2bc-2ab+2ac+a²+b²+c²-2ac-2bc+2ab

=4a²+4b²+4c²+4ab+4bc

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

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

=4

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

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

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

\(=x^6-1\)

Bài 1:

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

\(=9x^2-6x+1-2x^2+x-6x+3\)

\(=7x^2-11x+4\)

a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)

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

=2x

b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)

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

\(=\dfrac{x+1}{3x}\)

c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)

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

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

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

a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )

b,\(\dfrac{x^2+2x+1}{3x^2+3x}=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{x+1}{3x}\) ( đk : \(x\ne-1\) )

c

=

Bài 1 :

(a^2+b^2)(x^2+y^2)=(ax+by)^2 
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2 
<=> a^2y^2 + b^2x^2 = 2abxy 
<=> a^2y^2 + b^2x^2 - 2abxy = 0 
<=> (ay - bx)^2 = 0 
=> ay - bx = 0 
=> ay = bx 
=> a/x = b/y ( x,y khác 0)

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