a) \(\left(x-5\right)\left(x+8\right)-\left(x+4\right)\left(x-1\right)\)

\(=\left(x^2+3x-40\right)-\left(x^2+3x-4\right)\)

\(=x^2+3x-40-x^2-3x+4\)

\(=-36\)

b)\(x^4\left(x^2-1\right)\left(x^2+1\right)\)

\(=x^4\left(x^4-1\right)\)

\(=x^8-x^4\)

phép đặt trên thực ra là chuẩn hóa bdt

:). Sử dụng Bất đẳng thức Schur.

Giải:

Đặt: \(a+b+c=p\)

       \(abc=r\)

       \(ab+bc+ac=q\)

Theo bất đẳng thức Schur:

=> \(p^2\ge3q\) , \(2p^3+9r\ge7pq\) => \(p^3-4pq+9r\ge0\)=> \(p^3-4pq+9\left(4-p\right)\ge0\Leftrightarrow p^3-4pq-9p+36\ge0\)(1)

và \(p^3\ge27r\)

Từ giả thiết ta có: \(p+r=4\)=> \(p^3+27\ge27r+27p=27\left(r+p\right)=27.4\)

=> \(p^3+27p-27.4\ge0\)\(\Leftrightarrow\left(p^3-27\right)+\left(27p-27.3\right)\ge0\)

\(\Leftrightarrow\left(p-3\right)\left(p^2+3p+9+27\right)\ge0\Leftrightarrow\left(p-3\right)\left(p^2+3p+36\right)\ge0\Leftrightarrow p-3\ge0\)

\(\Leftrightarrow p\ge3\)

Vì a, b, c >0 => \(abc>0\)=> r>0

=> \(3\le p< 4\)

=> \(\left(p+3\right)\left(p-4\right)\left(p-3\right)\le0\Leftrightarrow p^3-4p^2-9p+36\le0\) (2)

Từ (1), (2) => \(-4pq\ge-4p^2\Leftrightarrow q\le p\) hay  ab+bc+ac\(\le\)a+b+c

"=" xảy ra : \(a=b=c\)

  và \(a+b+c+abc=4\)

<=> a=b=c=1

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right):2}=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right)}\times\frac{1}{2}=\frac{2011}{2013}\)

\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4016}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

Vậy x = 2012

Chúc mừng bạn Xyz nha

Chúc bạn luôn học giỏi

1) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}=\frac{2x-4y+3z}{2.2-4.3+3.7}=\frac{-39}{13}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.2=-6\\y=-3.3=-9\\z=-3.7=-21\end{cases}}\)

2) \(9x=10y\Leftrightarrow\frac{x}{10}=\frac{y}{9},4y=3z\Leftrightarrow\frac{y}{9}=\frac{z}{12}\)

suy ra \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Leftrightarrow\hept{\begin{cases}x=6.10=60\\y=6.9=54\\z=6.12=72\end{cases}}\)

3) \(3x=4y=6z\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x-y+z}{4-3+2}=\frac{-9}{3}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.4=-12\\y=-3.3=-9\\z=-3.2=-6\end{cases}}\)

\(\Leftrightarrow a.ab=111\Rightarrow a=3;b=7\)=>ab=37

 a x ab x b = bbb

4 x 44 x 4 = 704

1,3,6,10,15,21,....... 

   Ta thấy :

3=1=2

6=3+3

10=6+4

15=10+5

21=15+6

         Vậy số cần tìm là: 21+7=28

Câu 1 cứ bình phương lên mà viết thôi

Câu 2:

(1 + 3 + 5 +...+ 97 + 99).(45.3) - (5.9.2 - 15.3) 

= {50.[2.1 + (50 - 1).2]} : 2.(45.3) - (5.9.2 - 15.3)

= 2500.135 - 45 

= 337455