Có: \(\left(x+y+z\right)^2-x^2-y^2-z^2\) 

\(=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2\)

\(=2xy+2yz+2xz\)

\(=2\left(xy+yz+xz\right)\)

\(\left[\left(x+y\right)+z\right]^2=\left[\left(x+y\right)^2+2.\left(x+y\right)z+z^2\right]=x^2+2xy+y^2+2xz+2yz+z^2\)\(+z^2\)

Thay vào: x^2+y^2+z^2+ 2xy+2yz+2xz - x^2 - y^2 - z^2= 2(xy+yz+xz) (đpcm)

52.143 -52.39 -8.26 =52.143-52.39-4.52 =52.(143-39-4) =52. 100 =5200 chúc bạn thành công !

52 . 143 - 52 . 39 - 8 . 26

= 52 . 143 - 52 . 39 - 4 . 52

= 52 . (143 - 39 -  4)

= 52 . 100  = 5200

toan la bai lop 789

a³ ( c - b² ) + b³ ( a - c² ) + c³ ( b - a² ) + abc ( abc - 1 )

= a³c - a³b² + b³a - b³c² + c³b - c³a² + a²b²c² - abc

= a²b²c² - b³c² - ( a²c³ - bc³ ) - ( a³b² - ab³ ) + ( a³c - abc )

= b²c² . ( a² - b ) - c³ ( a² - b ) - ab² ( a² - b ) + ac ( a² - b ) 

= ( a² - b ) ( b²c² - c³ - ab² + ac )

= ( a² - b ) ( b² - c ) ( c² - a )

hình như (x^2+2)^2+1

sai rồi chị ak

a) \(A=\left(x^2-10x+25\right)\)\(-28\)

   \(A=\left(x-5\right)^2-28\)\(>=\)-28

MinA = -28 <=> x-5=0 <=> x=5

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

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

MaxB = 6 <=> x+1=0 <=> x=-1

c)\(C=-5\left(x^2-\frac{6}{5}x+\frac{9}{25}\right)-\frac{26}{5}\)

   \(C=-5\left(x-\frac{3}{5}\right)^2-\frac{26}{5}\)\(< =-\frac{26}{5}\)

MaxC = \(-\frac{26}{5}\)<=> \(x-\frac{3}{5}=0\)<=> x=\(\frac{3}{5}\)

d)\(D=-3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)+\frac{61}{12}\)

\(D=-3\left(x+\frac{1}{6}\right)^2+\frac{61}{12}\)\(< =\frac{61}{12}\)

MacD = \(\frac{61}{12}\)<=> \(x+\frac{1}{6}=0\)<=> \(x=\frac{-1}{6}\)

Đúng thì nhớ tích cho minh nha

a³(c - b²) + b³(a - c²) + c³(b - a²) + abc(abc - 1)

= a³c - a³b² + ab³ - b³c² + bc³ - a²c³ + a²b²c² - abc

= a²b²c² - b³c² - (a²c³ - bc³) - (a³b² - ab³) + (a³c - abc)

= b²c²(a² - b) - c³(a² - b) - ab²(a² - b) + ac(a² - b)

= (a² - b)(b²c² - c³ - ab² + ac) = (a² - b)[c²(b² - c) - a(b² - c)] = (a² - b)(b² - c)(c² - a)