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\(\left(-1-\frac{1}{12}\right).\left(-1-\frac{1}{13}\right).\left(-1-\frac{1}{14}\right)...\left(-1-\frac{1}{2017}\right)\)
\(=\frac{-13}{12}.\frac{-14}{13}.\frac{-15}{14}...\frac{-2018}{2017}\)
\(=\frac{-13}{12}.\frac{14}{-13}.\frac{-15}{14}...\frac{2018}{-2017}\)
\(=\frac{\left(-13\right).14.\left(-15\right)...2018}{12.\left(-13\right).14...2017}=\frac{2018}{12}=\frac{1009}{6}\)
a.
\(\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)\)
\(=\left(2+1\right)+\left(\frac{5}{6}+\frac{4}{9}\right):\left(10-9\right)+\left(\frac{1}{12}-\frac{1}{2}\right)\)
\(=3+\left(\frac{15}{18}+\frac{8}{18}\right):1+\left(\frac{1}{12}-\frac{6}{12}\right)\)
\(=\left(3+\frac{23}{18}\right):\left(1+\frac{-5}{12}\right)\)
\(=\frac{77}{18}:\frac{7}{12}\)
\(=\frac{77}{18}.\frac{12}{7}=\frac{22}{3}\)
Chúc bạn học tốt!!!
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)
a . ( -1/3 ) . 9/11 + ( -8/9 ) . 27
= 9/3 . ( -1/11 ) + 27/9 . ( -8 )
= 3 . ( -1/11 ) + 3 . ( -8 )
= 3 . ( -1/11 + ( -8 ) )
= 3 . ( -89/11 )
= -267/11
b . ( 1/2 - 13/14 ) : 5/7 - ( - - 2/21 + 1/7 ) : 5/7
= -3/7 : 5/7 - 5/21 : 5/7
= ( -3/7 - 5/21 ) : 5/7
= -2/3 : 5/7
= -14/15
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\) _ \(\frac{3-\frac{3}{11}-\frac{3}{17}}{5-\frac{5}{11}-\frac{5}{17}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)_ \(\frac{3\left(1-\frac{1}{11}-\frac{1}{17}\right)}{5\left(1-\frac{1}{11}-\frac{1}{17}\right)}\)= \(\frac{2}{4}-\frac{3}{5}\)= \(\frac{-1}{10}\)
\(\frac{1}{2007}.\left(\frac{1001}{2006}-2007\right)-\left(\frac{1}{2006}-2007\right).\frac{1001}{2007}\)
\(=\left(\frac{1001}{2007.2006}-\frac{2007}{2007}\right)-\left(\frac{1001}{2006.2007}-\frac{2007.1001}{2007}\right)\)
\(=\frac{1001}{2007.2006}-\frac{1001}{2006.2007}-1+1001\)
\(=-1+1001\)
\(=1000\)
\(1001\left(\frac{2}{12012}+\frac{74}{555555}-\frac{2}{13.7.11.5}\right)=\frac{1001.2}{12012}+\frac{1001.74}{555555}-\frac{1001.2}{13.11.7.5}=\frac{1001.2}{12.1001}+\frac{1001.2.37}{1001.37.15}-\frac{1001.2}{1001.5}\)\(=\frac{1}{6}+\frac{2}{15}-\frac{2}{5}=\frac{5}{30}+\frac{4}{30}-\frac{12}{30}=-\frac{3}{30}=-\frac{1}{10}\)
\(1001\left(\frac{2}{12012}+\frac{74}{555555}-\frac{2}{13.7.77.5}\right)=\frac{1001.2}{12.1001}+\frac{1001.74}{555.1001}-\frac{1001.2}{85.1001}=\frac{1}{6}+\frac{74}{555}-\frac{2}{85}=11877\)