Trả lời:

(tớ sửa lại nhé)

x⁴y⁴ + 4

= x⁴y⁴ + 4 + 4x²y² - 4x²y² 

= ( x⁴y⁴ + 4x²y² + 4 ) - 4x²y² 

= [ (x²y²)² + 2.x²y².2 + 2² ] - 4x²y² 

= ( x²y² + 2 )² - (2xy)² 

= ( x²y² + 2 - 2xy )( x²y² + 2 + 2xy )

Trả lời:

x⁴y⁴ + 4 

= x⁴y⁴ + 4 + 4x²y² - 4x²y² 

= ( x⁴y⁴ + 4x²y² + 4 ) - 4x²y² 

= [ ( xy )² + 2.xy.2 + 2² ] - ( 2xy )² 

= [ ( xy )² + 2 ]² - ( 2xy )²

= ( x²y² + 2 )² - ( 2xy )² 

= ( x²y² + 2 - 2xy )( x²y² + 2 + 2xy )

2(a² + b²) = (a+b)²

2a² + 2b² = a² + 2ab + b²

=> 2a^2 + 2b^2 - a^2 - 2ab - b^2 = 0

=> a^2 + b^2 - 2ab = 0

=> (a-b)^2 = 0

=> a - b = 0

=> a = b

a, bạn xem lại đề 

b, \(\frac{x^3-1}{x^2-1}=\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+x+1}{x+1}\)

Thay x = 6 ta được : \(=\frac{36+6+1}{6+1}=\frac{43}{7}\)

c, \(\frac{x^2-2x+1}{x^3-1}+\frac{x^2-1}{\left(x-1\right)^2}=\frac{x-1}{x^2+x+1}+\frac{x+1}{x-1}\)

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

giúp mik vs mik cần gấp

a) A = 1002 - 992 + 982 - 972 + ..... + 22 - 12 

       = ( 1002 - 992 ) + ( 982 - 972 ) + ..... + ( 22 - 12 ) 

     Số nhóm của dãy là : 

       [ ( 1002 - 12 ) : 10 + 1 ] : 2 = 50 nhóm 

       =      10              +      10        + .........      +     10 

      =               10 x 50 

       =               500

Hok tốt!!!!!!!!!!

\(a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=\left(a^2+b^2+c^2\right)^2\)

Làm đầy đủ giúp mình nhé

B = (x³ - y³) + (x - y)² 

 = (x - y)³ + 3xy(x - y) + (x - y)²

= (x - y)[(x - y)² + 3xy + (x - y)] 

= 4.(4² + 3.5 + 4) 

= 4.35 = 140

32×32

\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

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

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

Ta có x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 

 = x⁶(x + 1) + x⁴(x + 1) + x²(x + 1) + (x + 1) 

 = (x + 1)(x⁶ + x⁴ + x² + 1) 

 = (x + 1)(x⁴ + 1)(x² + 1) 

Khi đó \(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}=\frac{\left(x+1\right)\left(x^4+1\right)\left(x^2+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left(x^4+1\right)\left(x^2+1\right)}{x-1}\)

Trả lời:

B = 72² + 144.16 + 16² - 12² 

B = 72² + 2.72.16 + 16² - 12² 

B = ( 72 + 16 )² - 12² 

B = ( 72 + 16 - 12 )( 72 + 16 + 12 )

B = 76.100

B = 7600

(12 × 18) : 3 = 12 :?× 18 =?× 18 =?

\(\sqrt{\left(\sqrt{7}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

\(=\left|3-\sqrt{7}\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|3-\sqrt{7}\right|+\left|\sqrt{3}-1\right|\)

\(=3-\sqrt{7}+\sqrt{3}-1=2-\sqrt{7}+\sqrt{3}\)

\(\sqrt{\left(\sqrt{7}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

\(=\left|\sqrt{7}-3\right|+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}\)

\(=\left|\sqrt{7}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left|\sqrt{7}-3\right|+\left|\sqrt{3}-1\right|\)

\(=3-\sqrt{7}+\sqrt{3}-1\) (vì \(\sqrt{7}-3< 0;\sqrt{3}-1>0\))

\(=2-\sqrt{7}+\sqrt{3}\)