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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
\(1,=38-42+14-25+27+15=27\)
\(2,=12+21-23+21-10=21\)
\(3,=57-725-605+53=-1220\)
\(4,=55+45+15-15+55-45=110\)
a/ =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
Bài 1:
1: =15+37+52-37-17=52-2=50
2: =38-42+14-25+27+15=62-42+29=20+29=49
Bài 1: Bỏ ngoặc rồi tính
3) (21-32) - (-12+32)=21-32-(-12)-32=21-32+12-32=-31
4) (12+21) - (23-21+10)=12+21-23+21-10=21
5) (57-725) - (605-53)=57-725-605+53=-1220
6) (55+45+15) - (15-55+45)=55+45+15-15+55-45=55+55=110
Bài 2: Tính các tổng sau một cách hợp lí
1) (-37) + 14 + 26 + 37=(-37+37)+(14+26)=0+40=40
2) (-24) +6 + 10 + 24=(-24+24)+(6+10)=0+16=16
3) 15 + 23 + (-25) + (-23)=(15-25)+(23-23)=-10+0=-10
4) 60 + 33 + (-50) + (-33)=(60-50)+(33-33)=10+0=10
5) (-16) + (-209) + (-14) + 209=(-16-14)+(-209+209)=-30+0=-30
6) (-12) + (-13) + 36 + (-11)=(-11-12-13)+36=-36+36=0
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 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)