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1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
a/ \(\frac{15}{-50}=-\frac{3}{10}\)
\(\frac{7}{10}\)
\(\frac{24}{-20}=-\frac{12}{10}\)
\(\Rightarrow-\frac{12}{10};-\frac{3}{10};\frac{7}{10}\)
KL: ..........................................
b/ \(\frac{17}{39}=\frac{17.20}{39.20}=\frac{340}{780}\)
\(\frac{11}{65}=\frac{11.12}{65.12}=\frac{132}{780}\)
\(\frac{9}{52}=\frac{9.15}{52.15}=\frac{135}{780}\)
\(\Rightarrow\frac{132}{780};\frac{135}{780};\frac{340}{780}\)
KL:.............................................
c/ \(\frac{17}{20}=\frac{17.9}{20.9}=\frac{153}{180}\)
\(\frac{-19}{30}=-\frac{19.6}{30.6}=-\frac{114}{180}\)
\(\frac{38}{45}=\frac{38.4}{45.4}=\frac{152}{180}\)
\(\frac{-13}{18}=-\frac{13.10}{18.10}=-\frac{130}{180}\)
\(\Rightarrow-\frac{130}{180};-\frac{114}{180};\frac{152}{180};\frac{153}{180}\)
KL:..................................
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
Bài1
a) 25/42 - 20/63 =5/18
b) 9/50 - 13/75 - 1/6 = -4/25
c) 2/15 - 2/65 - 4/39 = 0
Bài2
a) x + 7/12 =17/18-1/9 b) 29/30 - (18/23 + x)=7/69
x + 7/12 = 5/6 18/23 + x =29/30 - 7/69
x =5/6 - 7/12 18/23 +x = 199/230
x = 1/4 x = 199/230 - 18/23
x= 19/230
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)