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)

bằng 1

\(\frac{254.399-145}{254+399.253}\)

\(=\frac{253.399+399-145}{254+399.253}\)

\(=\frac{253.399+253}{254+399.253}\)

\(=1\)

\(\Rightarrow\)

\(\text{Trả lời : }\)

\(\frac{254\cdot399-145}{254+399\cdot253}=\frac{399-145}{254}=\frac{254}{254}=1\)

\(\text{Chúc bạn học tốt ! }\)

Bạn Phúc làm đúng rồi càn bạn kia lam sai

399x253+254/399+(253+1)-145

=399x253+254/399x253+399-145

=399x253+254/399x253+254

=1

Kb và cho mik nha mọi người 

😊😊😊😊😊😊😊😊😊

cho mik nha

😍😍😍😍😍😍😍😍

#)Giải :

A, \(\frac{254x399-145}{254+399x253}\)

\(=\frac{253x399+399-145}{254+399x253}\)

\(=\frac{253x399+254}{254+399x253}\)

\(=1\)

B, \(\frac{5932+6001x5931}{5931x6001-69}\)

\(=\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)

\(=\frac{5932+6001x5931}{5931x6001+6001-69}\)

\(=\frac{5932+6001x5932}{5932x6001+5932}\)

\(=1\)

         #~Will~be~Pens~#

\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)

\(=1\)

\(\frac{254x399-145}{254+399x253}\)

=\(\frac{\left(253+1\right)x399-145}{254+399x253}\)

=\(\frac{253x399+399-145}{254+399x253}\)

=\(\frac{253.399+254}{254+399x253}\)=\(1\)

\(\frac{5932+6001x5931}{5931x6001-69}\)

=\(\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)

=\(\frac{5932+6001x5931}{5931x6001+6001-69}\)

=\(\frac{5932+6001x5931}{5931x6001x5932}\)=1

# Hok tốt

a)  \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)

    \(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)

C) bạn chỉ ần bỏ các số giống nhau thôi nhé

   = 1

b) 

Do con lợn TNT học giỏi không biết làm phần b ) .  , Trắng sẽ giúp bạn :

1/2 + 1/6 + 1/12 + ...+ 1/110

= 1/1.2 + 1/2. 3 + 1/3 . 4 + ... + 1/10 . 11 

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/10 - 1/11

= 1 - 1/11 

= 10/11

a, \(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}.\frac{4}{15}\)

\(=\frac{16}{45}\)

Bạn kia làm câu a sai rồi

C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)

C= CÂU HỎI TƯƠNG TỰ 

=> đcpm 

\(A=\frac{254\cdot399-145}{254+399\cdot253}\)

\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)

\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)

\(A=\frac{253\cdot399+254}{254+399\cdot253}\)

\(A=1\)

\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)

\(B=1\)

\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)

\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)

\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)

\(C=\frac{1}{2017}\)

2: \(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left[\dfrac{7}{3}\cdot\left(4+\dfrac{1}{2}-2\right)\right]\cdot\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{4}{3}\cdot\dfrac{5}{2}=\dfrac{59}{15}-\dfrac{10}{3}=\dfrac{59-50}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)

4: \(=254+254\cdot399+399\)

\(=254\cdot400+399\)

\(=101600+399=101999\)

5: \(=638+3900-41=3900+597=4497\)