Thay a; b vào biểu thức ta có : 

2475 x 5 + 1968 : 4

= 12375 + 492

= 12867

Chúc bạn học tốt !!!!

thay ta có\(2475.5+1968:4\)

\(\Leftrightarrow12375+492\)

\(=12867\)

ab.cc.abc = abc.1001  

ab.cc = 1001  

ab.c.11 = 11.91  

ab.c = 91 = 13 x 7  

Hay a=1 ; b=3 ; c=7  

Số abc = 137    

Thử lại:  13x77x137 = 137137

Ta có :

\(x+y=a\)\(\Rightarrow\left(x+y\right)^2=a^2\)

\(\Rightarrow x^2+y^2+2xy=a^2\)

\(\Rightarrow2xy=a^2-b\)\(\Leftrightarrow xy=\frac{a^2-b}{2}\)

Thay vào ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=a\left(b-\frac{a^2-b}{2}\right)\)

\(.....\)

TL:

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

Thay x+y=a; \(x^2+y^2=b\) Ta đc:

\(a.\left(b-xy\right)=ab-axy\) 

Vậy \(x^3+y^3=ab-axy\) 

hc tốt

\(\hept{\begin{cases}x+y=2\\x^2+y^2=10\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2+2xy=2\\x^2+y^2=10\end{cases}\Leftrightarrow}\hept{\begin{cases}xy=-4\\x^2+y^2=10\end{cases}}\)

Có \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=2\left(10+4\right)=28\)

\(x+y=2\Rightarrow\left(x+y\right)^2=4\)

\(\Rightarrow x^2+2xy+y^2=4\)

\(\Rightarrow10+2xy=4\)\(\Leftrightarrow2xy=-6\Rightarrow xy=-3\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10-\left(-3\right)\right)=26\)

#)Giải :

Câu 1 :

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=1-\frac{2}{x+1}=\frac{2009}{2011}\Rightarrow x=2010\)

\(\hept{\begin{cases}x+y=m\\x-y=n\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2+2xy=m^2\\x^2+y^2-2xy=n^2\end{cases}\Leftrightarrow}4xy=m^2-n^2\Leftrightarrow xy=\frac{m^2-n^2}{4}\)

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

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

                            \(=n\left(m^2-\frac{m^2-n^2}{4}\right)\)

Rut gon not

Áp dụng bđt côsi cho 3 số ta được \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=3\)

\(\Rightarrowđpcm\)

\(x^4+2x^3-16x^2-2x+15\)

\(=x^4+5x^3-3x^3-15x^2-x^2-5x+3x+15\)

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

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

\(=\left(x+5\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]\)

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

\(=\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+5\right)\)