2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7

ngu

Đặt 5x=4y=2z=k suy ra \(x=\frac{k}{5};y=\frac{k}{4};z=\frac{k}{2}\)

Ta có : 

x-y+z=-18

\(\frac{k}{5}-\frac{k}{4}+\frac{k}{2}=-18\)

\(k.\left(\frac{1}{5}-\frac{1}{4}+\frac{1}{2}\right)=-18\)

\(k.\frac{9}{20}=-18\)

k = -40 suy ra x = -8 ; y = -10 ; z = -20

Ta có:

\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}=\left(\frac{2}{-8}+\frac{5}{-40}+\frac{5}{-20}\right)^{2016}=\left(\frac{5}{-8}\right)^{2016}=0\)

a,A=5x²z-10xyz+5y²z

=5z(x²-2xy+y²)

=5z(x-y)²

Thay x=124,y=24,z=2 vào A ta được:

A=5.2(124-24)²=10.100²=10000

b,B=2x²+2y²-x²z+z-y²z-2

=2(x²+y²)-z(x²+y²)+(z-2)

=(2-z)(x²+y²)-(2-z)

=(2-z)(x²+y²-1)

Thay x=1,y=1,z=-1 vào B

B=(2+1)(1²+1²-1)=3

c, C=x²-y²+2y-1

=x²-(y²-2y+1)

=x²-(y-1)²

=(x-y+1)(x+y-1)

=(75-26+1)(75+26-1)

=50.100=5000

Thay x = 4 vào A ta được:

5.4⁵ - 5.4⁴ + 5.4³ - 5.4² + 5.4 - 1

= 5.1024 - 5.256 + 5.64 - 5.16 + 5.4 - 1

= 5120 - 1280 + 320 - 80 + 20 - 1

= 4099

Kq = 4099 nha

( x - 1 )²⁰¹⁸ + (y - 2 )²⁰²⁰+(z-3)²⁰²²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

\(A=\dfrac{1}{9}\left(-x\right)^{2021}y^2z^3=\dfrac{1}{3}\left(-1\right)^{2021}.2^2.3^3=\dfrac{1}{3}.\left(-1\right).4.27=-36\)

a: A=y(x-4)-5(x-4)

=(x-4)(y-5)

Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5

b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)

=0,2*10=2

d: Khi x=5,75 và y=4,25 thì

D=5,75^3-5,75^2*4,25+4,25^3

=8087/64

x=4

=>x+1=5

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

  =x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1

  =x^6-x-1

  =4^6-4-1

  =4091

\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)

\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)

b)tương tự

=x^2006-x^2006-x^2005+x^2005+x^2004-...+x^3-x^2-x^2-x-5

=-x-5

=-7-5=-12

Bài 1 : 

a, \(\left(2x^2-3x-1\right)\left(5x+2\right)=10x^3+4x^2-15x^2-6x-5x-2\)

\(=10x^3-11x^2-11x-2\)

b, sửa đề :  \(\left(-x^2+2x-3\right)\left(4x^2-2x+3\right)\)

\(=-4x^4+2x^3-3x^2+8x^3-4x^2+6x-12x^2+6x-9\)

\(=-4x^4+10x^3-19x^2+12x-9\)

Bài 2 : 

\(B=\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\)

Thay x = 1 ; y = 1 ; z = -1 vào biểu thức trên ta được 

\(B=\left(1+1\right)\left(-2+1\right)+\left(1-1\right)\left(y-z\right)=2.\left(-1\right)=-2\)

Trả lời:

Bài 1: 

a, ( 2x² - 3x - 1 ) ( 5x + 2 ) 

= 10x³ + 4x² - 15x² - 6x - 5x - 2

= 10x³ - 11x² - 11x - 2

b, ( - x² + 2x - 3 ) ( 4x² - 2 + 3 )

= - 4x⁴ - 2x² + 3x² + 8x³ - 4x + 6x - 12x² + 6 - 9

= - 4x⁴ + 8x³ - 11x² + 2x - 3

Bài 2:

B = ( 2x + y ) ( 2z + y ) + ( x - y ) ( y - z ) 

Thay x = 1, y = 1, z = - 1 vào B, ta được:

B = ( 2.1 + 1 ) [ 2.( - 1 ) + 1 ] + ( 1 - 1 ) [ 1 -  ( - 1 )

= ( 2 + 1 ) ( - 2 + 1 ) + 0 . ( 1 + 1 )

= 3 . ( - 1 ) + 0

= - 3