x<y suy ra a/m<b/m suy ra a<b (vì m<0)

mà a<b suy ra a+b < b+b

suy ra a+b<2b

suy ra a+b/2 <b

suy ra a+b/2m <b/m

suy ra a+b/2m< y

Suy ra z<y   (1)

Mặt khác a<b suy ra a+a <a+b

suy ra 2a <a+b

suy ra 2a/m <a+b/ m

suy ra a/m < a+b/2m

suy ra x<z    (2)

Từ (1) và (2)

suy ra x<z<y

a^3 + b^3 + c^3 - 3abc

=(a^3+3a^2b+3ab^2+b^3)+c^3-(3a^2b+3ab^2+3abc)

=(a+b)^3+c^3-3ab(a+b+c)

=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)

=(a+b+c)(a^2+b^2+c^2+2ab-3ab-bc-ac)

=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)

Thay a + b + c = 0, ta có:

0(a^2+b^2+c^2-ab-bc-ac)

=0

Vậy nếu a + b + c = 0 thì a^3 + b^3 + c^3 - 3abc = 0

Theo bài ra, ta có: a+b+c
Suy ra: 3(a+b+c)-3abc=0
Suy ra: -3abc=0
Tương đương: -3*(b+c)*(a+c)*(a+b)=0
Tương đương: -3* a^2+b^2+c^2=0
Tương đương: -3*0=0
Suy ra: nếu a+b+c=0 thì a3+b3+c3-3abc=0(đpcm)

a, 0

b,0

c, 0

mình ko chắc lắm

a/ (x+y)(x+y)

   =x+y.x+y

   =x+x.y+y

   =2.x.2.y

    =2.(x+y)

\(A=\left(\frac{4x}{x^2-4}+\frac{2x-4}{x+2}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\=\left(\frac{4x}{x^2-4}+\frac{\left(2x-4\right)\left(x-2\right)}{x^2-4}\right)\frac{x+2}{2x}+\frac{2}{2-x}=\left(\frac{4x}{x^2-4}+\frac{2x^2-4x-4x+8}{x^2-4}\right) \frac{x+2}{2x}+\frac{2}{2-x}\)

\(=\left(\frac{4x+2x^2-8x+8}{x^2-4}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\ =\frac{2x\left(x+2\right)-8\left(x-1\right)}{x^2-4}.\frac{x+2}{2x}+\frac{2}{2-x}\)

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

\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)

\(\Leftrightarrow5x=-17\)

\(\Rightarrow x=-\frac{17}{5}\)

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

\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)

\(\Leftrightarrow10=1\)

=> vô nghiệm 

c) \(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\)

\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+8x+8-22x-27\)

\(\Leftrightarrow8x=-24\)

\(\Rightarrow x=-3\)

a) 2( x - 1 )² + ( x + 3 )² = 3( x - 2 )( x + 1 )

<=> 2( x² - 2x + 1 ) + x² + 6x + 9 = 3( x² - x - 2 )

<=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

<=> 2x² - 4x + x² + 6x - 3x² + 3x = -6 - 2 - 9

<=> 5x = -17

<=> x = -17/5

b) ( x + 2 )² - 2( x - 3 ) = ( x + 1 )²

<=> x² + 4x + 4 - 2x + 6 = x² + 2x + 1

<=> x² + 4x - 2x - x² - 2x = 1 - 4 - 6

<=> 0x = -9 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x - 1 )² + ( x - 2 )² = 2( x + 4 )² - ( 22x + 27 )

<=> x² - 2x + 1 + x² - 4x + 4 = 2( x² + 8x + 16 ) - 22x - 27

<=> 2x² - 6x + 5 = 2x² + 16x + 32 - 22x - 27

<=> 2x² - 6x - 2x² - 16x + 22x = 32 - 27 - 5

<=> 0x = 0 ( đúng ∀ x ∈ R )

Vậy phương trình nghiệm đúng ∀ x ∈ R

a) 2(x - 1)² + (x + 3)² = 3(x - 2)(x + 1)

=> 2(x² - 2x + 1) + x² + 6x + 9 = 3(x² - x - 2)

=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

=>3x² + 2x + 11 = 3x² - 3x - 6

=> 3x² + 2x + 11 - 3x² + 3x + 6 = 0

=> 5x  + 17 = 0

=> 5x = -17

=> x = -17/5

b) (x + 2)² - 2(x - 3) = (x + 1)²

=> x² + 4x + 4 - 2x + 6 = x² + 2x + 1

=> x² + 4x + 4 - 2x + 6 - x² - 2x - 1 = 0

=> (x² - x²) + (4x - 2x - 2x) + (4 + 6 - 1) = 0

=> 9 = 0(vô lí)

c) (x - 1)² + (x - 2)² = 2(x + 4)² - (22x + 27)

=> x² - 2x + 1 + x² - 4x + 4 = 2(x² + 8x + 16) -22x - 27

=> x² - 2x + 1 + x² - 4x + 4 = 2x² + 16x + 32 - 22x - 27

=> (x² + x²) + (-2x - 4x) + (1 + 4) = 2x² + (16x - 22x) + (32 - 27)

=> 2x² - 6x + 5 = 2x² - 6x + 5

=> 2x² - 6x + 5 - 2x² + 6x - 5 = 0

=> (2x² - 2x²) + (-6x + 6x) + (5 - 5) = 0

=> 0 = 0(đúng)

a: Để A là số nguyên thì \(x^3-3x^2-x^2+3x+x-3-7⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{4;2;10;-4\right\}\)

b: Đề sai rồi bạn

Ta có : a ≡ b (mod m) => a - b
m (m (a - b)
và b ≡ c (mod m) => b - c
m (m (b - c)
Vì a - c = (a - b) + (b - c) => a - c
m (tính chất chia hết của tổng) hay 
a ≡ c (mod m).

Bài 1:

a) \(\frac{x-1}{0-2}=\frac{1,2}{1,5}\)

\(\Leftrightarrow\frac{1-x}{2}=\frac{4}{5}\)

\(\Leftrightarrow5-5x=8\)

\(\Leftrightarrow x=-\frac{3}{5}\)

b) Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y+2z}{4-6+6}=\frac{16}{4}=4\)

\(\Rightarrow\hept{\begin{cases}x=4\\y=8\\z=12\end{cases}}\)

Bài 1:

c) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)

\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)

d) \(x:y:z=3:5:2\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{5x-7y+5z}{15-35+10}=\frac{124}{-10}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{186}{5}\\y=-62\\z=-\frac{124}{5}\end{cases}}\)

Bài 1:

e) \(\frac{x}{3}=\frac{y}{5}=k\left(k\inℝ\right)\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=5k\end{cases}}\)

Ta có: \(x\cdot y=240\Leftrightarrow15k^2=240\)

\(\Leftrightarrow k^2=16\Rightarrow k=\pm4\)

=> \(\hept{\begin{cases}x=\pm12\\y=\pm20\end{cases}}\)