\(\frac{2x+3}{x-5}\)\(=\frac{2\left(x-5\right)+13}{x-5}\)

                     \(=\frac{2\left(x-5\right)}{x-5}+\frac{13}{x-5}\)

                     \(=2+\frac{13}{x-5}\)

để biểu thức trên có giá trị nguyên <=> \(\frac{13}{x-5}\)thuộc Z

mà  \(x\)thuộc Z => \(x-5\)thuộc ước của \(13\)

=> \(x-5\)thuộc \(\left(1;-1;13;-13\right)\)

=>\(x\)thuộc \(\left(6;4;18;-8\right)\)

vậy ....

\(\frac{x^3-2x^2+4}{x-2}\) \(=\frac{x^2\left(x-2\right)+4}{x-2}\)

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

để biểu thức trên đạt giá trị nguyên <=> \(\frac{4}{x-2}\) thuộc giá trị nguyên

  mà \(x\) là số nguyên => \(x-2\)thuộc ước của \(4\)

=> \(x-2\) thuộc \(\left(1;-1;2;-2;4;-4\right)\)

=>   \(x\)thuộc \(\left(3;1;4;0;6;-2\right)\)

vậy...

a)x^2-4x+2=(x-4)²-2=(x-4-\(\sqrt{2}\))(x-4+\(\sqrt{2}\))

b)x^2+x+2=(x+\(\frac{1}{4}\))²+\(\frac{3}{4}\)

c)x^2-x+3=(x-\(\frac{1}{4}\))²+\(\frac{13}{4}\)

d)x^2-6x+5=(x-5)(x-1)

e)x^2-4x+5+(x-4)²+1

Ta có : 

\(A=10^3-3^3-7^3\)

\(A=10^3-\left(3^3+7^3\right)\)

\(A=10^3-\left(3+7\right)\left(3^2-3.7+7^2\right)\)

\(A=10^3-10\left(9-21+49\right)\)

\(A=10^3-10.37\)

\(A=10\left(10^2-37\right)\)

\(A=10\left(100-37\right)\)

\(A=10.63\)

\(A=630\)

Vậy \(A=630\)

Chúc bạn học tốt ~ 

A = 630

Mình chắc 100 %

Ta có : x + y = 2 

=> \(\left(x+y\right)^2=4\)

<=> x² + 2xy + y² = 4

=> 2xy + 10 = 4

=> 2xy = -6

=> xy = -6

P = x³ + y³ = (x + y)(x² - xy + y²) 

= 2(10 + 6) 

= 2.16

=32

x+y=2 

<=>(x+y)²=4

<=>x²+2xy+y²=4

<=>2xy+10=4

<=>2xy=-6

<=>xy=-3

Ta có: P=x³+y³=(x+y)(x²-xy+y²)=2(10+3)=2.13=26

\(A=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

\(A=x^3-y^3-3xy\)

\(\left(x-y\right)^3=\left(x^2-2xy+y^2\right)\left(x-y\right)=x^3-2x^2y+xy^2-x^2y+2xy^2-y^3\)

\(=x^3-3x^2y+3xy^2-y^3\)

\(=x^3-y^3-3\left(x^2y-xy^2\right)\)

\(=x^3-y^3-3xy\left(x-y\right)\)

\(=x^3-y^3-3xy.1=x^3-y^3-3xy\)

=> \(A=x^3-y^3-3xy=\left(x-y\right)^3=1^3=1\)

\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left[\left(a^3+b^3\right)+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)

Vì a+b+c=0 \(\hept{\begin{cases}a>0\\b>0\\c>0\end{cases}}\)

Do đó: \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow a=b=c}\)

Miyuki Misaki cm ngược rồi 
Ta có : a + b + c = 0 

<=> a + b = -c {...........}

<=> (a + b)³ = -c³

<=> a³ + b³ + 3ab(a + b) = -c³

<=> a³ + b³ + c³ = -3ab(a + b) 

<=> a³ + b³ + c³ = -3ab(-c) {vì a + b = -c}

<=>  a³ + b³ + c³ = 3abc

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)\)

\(=3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(đpcm\right)\)

Theo bài ra ta có:

\(\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Leftrightarrow\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

$(a+b+c{)}^3=((a+b)+c{)}^3=(a+b{)}^3+c^3+3(a+b\left)c\right(a+b+c)$
$=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)$
$=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c\left)\right)$
$=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)$
$=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$