\(x^{50}+x^{49}+x^{48}+...+x^2+x+1\)

\(=\left(x^{50}+x^{49}+...+x^{34}\right)+\left(x^{33}+x^{32}+...+x^{17}\right)+\left(x^{16}+x^{15}+..+1\right)\)

\(=x^{34}\left(x^{16}+x^{15}+...+1\right)+x^{17}\left(x^{16}+x^{15}+...+1\right)+\left(x^{16}+x^{15}+...+1\right)\)

\(=\left(x^{46}+x^{45}+...+1\right)\left(x^{34}+x^{17}+1\right)\)

Vậy...

\(36^2+72.64+64^2\)

\(=36^2+2.36.64+64^2\)

\(=\left(36+64\right)^2\)

\(=100^2\)

\(=10000\)

Đáp án:

 10000.10000.

Giải thích các bước giải:

36²+72.64+64².36²+72.64+64².

=36²+2.36.64+64².=36²+2.36.64+64².

=(36+64)².=(36+64)².

=100².=100².

=10000.

MTC: (x+y)(x+1)(1-y)

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

\(=x-y+xy\)

Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định

( 4x - 1 )³ - ( 4x - 3 )( 16x² + 3 )

= ( 4x )³ - 3.( 4x )².1 + 3.4x.1² - 1³ - ( 64x³ + 12x - 48x² - 9 )

= 64x³ - 48x² + 12x - 1 - 64x³ - 12x + 48x² + 9

= ( 64x³ - 64x³ ) + ( 48x² - 48x² ) + ( 12x - 12x ) + ( 9 - 1 )

= 8

a,\(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)

\(=8\)

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

ĐK : ...

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

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

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

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

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

Mình sửa dòng 5 một chút nhé 

\(=\frac{\left(x^2-2\right)\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)( như kia dễ bị nhầm )

( 2x + 3 )( 4x² - 6x + 9 ) - 2( 4x³ - 1 )

= ( 2x )³ + 3³ - 8x³ + 2

= 8x³ + 27 - 8x³ + 2

= 29

[2x+3][4x2L- 6X+9L] - 2L[4x3l -1]

=[2xl]3l + 3l3 - 8x3l +2

=8x3l+27-8x3l +2 

= 29ll= 29l = 29

( x + 3 )( x - 3 ) - ( x + 5 )( x - 1 ) - ( x - 4 )²

= x² - 9 - ( x² + 4x - 5 ) - ( x² - 8x + 16 )

= x² - 9 - x² - 4x + 5 - x² + 8x - 16

= -x² + 4x - 20

= x^2 - 3^2 - ( x^2 + 4x - 5 ) - ( x^2 - 8x + 16 ) 

= x^2 - 9 - x^2 - 4x + 5 - x^2 +8x - 16 

= -x^2 + 4x - 20