\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\frac{\left(a+b\right)^3}{c^3}+\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}\)

\(\frac{\left(a^2b+ab^2\right)^3+\left(b^2c+c^2b\right)^3+\left(c^2a+a^2c\right)^3}{\left(abc\right)^3}\)

\(\left(\frac{a}{c}+\frac{b}{c}\right)^3+\left(\frac{b}{a}+\frac{c}{a}\right)^3+\left(\frac{c}{b}+\frac{a}{b}\right)^3\)

\(\left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3\)

dễ thấy \(\frac{c}{a+b}=\frac{1}{2}< =>\frac{a+b}{c}=2\)

làm tương tự với 3 cái còn lại ta đc:

\(2^3+2^3+2^3=24\)

hello , kết bạn nhắn tin ko

a) (x + 5)² - x² = 45

<=> (x + 5 - x)(x + 5 + x) = 45

<=> 5(2x + 5) = 45

<=>> 2x + 5 = 9

<=> 2x = 4

<=> x = 2

Vậy S = {2}

b) (2x + 1)(1 - 2x) + (2x - 1)² = 22

<=> (2x - 1)(-2x - 1 + 2x - 1) = 22

<=> -2(2x - 1) = 22

<=> 2x - 1 = -11

<=> 2x = -10

<=> x = -5

c) (3x - 5)² - ((3x + 1)² = -3

<=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = -3

<=> -6(6x - 4) = -3

<=> 6x - 4 = 1/2

<=> 6x = 9/2

<=> x = 3/4

a,(x+5)²-x²=45

x²+10x+25-x²=45

10x+25=45

10x=20=> x=2

b,(2x+1)(1-2x)+(2x-1)²=22

(1+2x)(1-2x)+4x²-4x+1=22

1-4x²+4x²-4x+1=22

2-4x=22

4x=-20

x=-5

a) 3(x - 1)² - 3x(x - 5) = 1 <=> 3(x² - 2x + 1) - 3x² + 15x = 1

<=>> 3x² - 6x + 3 - 3x² + 15x = 1

<=> 9x = -2

<=> x = -2/9

b) 2x(8x - 3) - (4x - 3)² = 27

<=> 16x² - 6x - 16x² + 24x - 9 = 27

<=> 18x  = 36

<=> x = 2

\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

\(\Leftrightarrow3x^2-6x+3-3x^2+15x=1\)

\(\Leftrightarrow9x+3=1\)

\(\Leftrightarrow9x=-2\)

\(\Leftrightarrow x=-\frac{2}{9}\)

Vậy \(x=-\frac{2}{9}\)

-8x^2+5x+3=(8x+3)(-x+1)

Xét hiệu a³ + b³ - ( a + b ) ta có :

a³ + b³ - ( a + b ) = a³ + b³ - a - b = ( a³ - a ) + ( b³ - b ) = a( a² - 1 ) + b( b² - 1 ) = a( a - 1 )( a + 1 ) + b( b - 1 )( b + 1 ) 

Vì a,b nguyên nên a , a - 1 , a + 1 và b , b - 1 , b + 1 là 3 số nguyên liên tiếp

=> a( a - 1 )( a + 1 ) ⋮ 3 và b( b - 1 )( b + 1 ) ⋮ 3

=> a( a - 1 )( a + 1 ) + b( b - 1 )( b + 1 ) ⋮ 3 hay a³ + b³ - ( a + b ) ⋮ 3

mà a + b ⋮ 3 => a³ + b³ ⋮ 3 ( đpcm )

\(A=25x^2-10x+5=\left(5x-1\right)^2+4\ge4\)

flo giúp nốt câu kia đi :)

A=25x²-10x+5=25x²-10x+1+4=(5x-1)²+4

ta có: (5x-1)²\(\ge0\)

suy ra (5x-1)²+4\(\ge4\) suy ra GTNN là 4

B=t²+12t²-3=13t²-3

ta có: 13t²\(\ge0\)

suy ra 13t²-3\(\ge-3\)suy ra GTNN là -3

Đề là gì vậy ???

Thu gon

Câu 1:

a, (x-1).(x²-x+1)=x³-x²+x-x²+x-1=x³-2x²+2x-1

b, (2a+b).(2a-b)=4a²-b²

Câu 2:

a,(2x-1).(x²-5x+3)=2x³-10x²+6x-x²+5x-3

                            = 2x³-11x²+11x-3

b,(-x²+4x-1).(-x+4)=x³-4x²-4x²+16x+x-4

                              = x³-8x²+17x-4

c,(-2x-3).(x+4)+(-x+1)= -2x²-8x-3x-12-x+1

                                  = -2x²-12x-11

d,(-3).(x+4).(x-7)+2.(x-5).(x+1)=-3x²+9x+84+2x²-8x-10

                                                = x²+x-74

Câu 3:

a,5x²-(2x+1).(x-2)-x.(3x+3)+7=5x²-2x²-4x-x+2-3x²-3x+7

                                               = -5x+9

b,(5x-2).(x+1)-(x-3).5x+1-17.(x-2)=5x²+5x-2x-2-5x²+15x+1-17x+34

                                                     =x+33