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Bài làm
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)
Hình như bạn nhầm câu hỏi rồi đay.Mik đang bai toan về nhân hai ô nguyên mà

a/ =15+37+52-37-17
=(15+52)-17+(37-37)
=67-17+0
=50
b/=38-42+14-25+27+15
=-4+14+27-25+15
=10+27-40
=37-40
=-3
c/=-21+32+12-32
=-21+12+32-32
=11+0
=11
d/=-12-21+23-23+21-10
=-33+0+11
=-22
e/=57-725-605+53
=-668-605+53
=-1273+53
=-1220
f/=55+45+15-15+55-45
=55+55+45-45+15-15
=110+0+0
=110

\(1,=38-42+14-25+27+15=27\)
\(2,=12+21-23+21-10=21\)
\(3,=57-725-605+53=-1220\)
\(4,=55+45+15-15+55-45=110\)

a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)
\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)
\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)
b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)
\(=-\frac{5}{72}:\frac{3.7}{8.9}\)
\(=-\frac{5}{72}:\frac{7}{24}\)
\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)
a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)
b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)
c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)
\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)
\(=\frac{-23}{25}+\frac{-27}{25}\)
\(=\frac{-50}{25}=-2\)

Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)

ó ai giúp mk ko , ai giúp mk nhanh và đúng mk sẽ k cho người đó nhé , ( nhanh chứ ko mk đâng cần)
1.
a, \(\left(15+37\right)+\left(52-37-17\right)\)
\(=15+37+52-37-17\)
\(=50\)
b, \(\left(38-42+14\right)-\left(25-27-15\right)\)
\(=38-42+14-25+27+15\)
\(=27\)
c, \(-\left(21-32\right)-\left(-12+32\right)\)
\(=-21+32+12-32\)
\(=-9\)
d, \(-\left(12+21-23\right)-\left(-23-21+10\right)\)
\(=-12-21+23+23+21-10\)
\(=24\)
e, \(\left(57-725\right)-\left(605-53\right)\)
\(=57-725-605+53\)
\(=-1220\)
f, \(\left(55+45+15\right)-\left(15-55+45\right)\)
\(=55+45+15-15+55-45\)
\(=110\)
2.
c, \(25\left(-47\right)\left(-4\right)\)
\(=-100\left(-47\right)=4700\)
d, \(8\left(125-30000\right)\)
\(=8.125-8.30000\)
\(=1000.240000\)
\(=240000000\)
e, \(512\left(2-128\right)-128\left(-512\right)\)
\(=512\left(2-128\right)+128.512\)
\(=512\left(2-128+128\right)\)
\(=1024\)
3.
4 B(-5) trog đó cs cả số âm là: 10; -20; 35; -40
tất cả các Ư(15) là: \(\pm1;\pm3;\pm5;\pm15\)
LƯU Ý: đề hơi dài, cs sai sót j mong e bỏ qua mà k cho cj nha