a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

ko hiểu

BIU BIU 

1, Thấy : \(\frac{1}{5}< \frac{2}{2.4}\)

                \(\frac{1}{13}< \frac{2}{4.6}\)

                  .....

                  \(\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2n\left(2n+1\right)}\)

Cộng từng vế có :

 \(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\)

\(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{4}+....+\frac{1}{2n}-\frac{1}{2n+2}\)

 \(\frac{1}{5}+\frac{1}{13}+..+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{2n+2}\)

Mà \(\frac{1}{2}-\frac{1}{2n+2}< \frac{1}{2}\)=> Tổng trên < 1/2

2,M = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

=> M \(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{\left(n-1\right)^2}-\frac{1}{n^2}+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

    \(M=1-\frac{1}{\left(n+1\right)^2}=\frac{\left(n+1\right)^2-1}{\left(n+1\right)^2}=\frac{n^2+2n+1-1}{\left(n+1\right)^2}=\frac{n^2+2n}{\left(n+1\right)^2}\)

Đến đây tắc r tự nghĩ tiếp >:

tại xao 1/n^2+(n+1)^2 < 2n(2n+1)?

ta co

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}>\frac{1}{2.2}+\frac{1}{3.3}+....+\frac{1}{10.10}\)

ma ve trai =\(1-\frac{1}{10}\)

nen ve phai <1

D=1/1.2+1/2.3+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

=1+0+0+0+...+0-1/10=1-1/10=9/10

ta có ; 1/2² +1/3²+...+1/2017²<1/1.2+1/2.3+1/3.4+.....+1/2016.2017=1-1/2+1/2-1/3+...+1/2016-1/2017=1+0+0+0+...+0-1/2017

=1-1/2017<1

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1\left(đpcm\right)\)

Ta có: S = \(\dfrac{1}{3}+\dfrac{3}{3.7}+\dfrac{5}{3.7.11}+...+\dfrac{2n+1}{3.7.11...\left(4n+3\right)}\)

⇒ 2S = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+2}{3.7.11...\left(4n+3\right)}\)

⇒ 2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+3}{3.7.11...\left(4n+3\right)}\)

Đến đây nó sẽ rút gọn liên tục và sau nhiều lần rút gọn ta có:

2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+\dfrac{1}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{11}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{1}{3.7}\) = \(\dfrac{2}{3}+\dfrac{7}{3.7}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

Suy ra 2S < 1 ⇒ S < \(\dfrac{1}{2}\)(đpcm)

1.Tính hợp lí:

-5/9.3/11+(-3/18).3/11

=3/11.[(-5/9)+(-3/18)]

=3/11.-13/18

=-39/198

=13/66

2.Tìm x biết:

2/3+7/4:x=5/6

       7/4:x=5/6-2/3

       7/4:x=1/6

             x=7/4:1/6

             x=21/2      

bài 1

\(\frac{-5}{9}.\frac{3}{11}+\frac{-3}{18}.\frac{3}{11}\)

= \(\frac{3}{11}.\left(\frac{-5}{9}+\frac{-3}{18}\right)\)

= \(\frac{-13}{66}\)

bài 2

\(\frac{2}{3}+\frac{7}{4}:x=\frac{5}{6}\)

\(\frac{7}{4}:x=\frac{1}{6}\)

\(x=\frac{21}{2}\)