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2) 1\26+1\27+1\28+........+1\50=1+1\2+1\3+......+1\50 -( 1+1\2+1\3+.....+1\25)=1+1\2+1\3+....+1\50-2.(1\2+1\4+1\6+....+1\50)=1-1\2+1\3-1\4+.....+1\49-1\50=vế phải(đpcm)
a)
\(\frac{8}{11}.\frac{14}{23}+\frac{9}{23}:\frac{11}{8}-\frac{8}{11} \)
\(=\frac{8}{11}.\frac{14}{23}+\frac{9}{23}.\frac{8}{11}-\frac{8}{11}\)
\(=\frac{8}{23}.(\frac{14}{23}+\frac{9}{23}-1)\)
\(=\frac{8}{23}.0\)
=0
b)
\(1,8.\frac{-20}{27}\)+(75%\(-\frac{5}{16}):3\frac{1}{2}\)
=\(\frac{9}{5}.\frac{-20}{27}+(\frac{3}{4}-\frac{5}{16}):\frac{7}{2}\)
=\(\frac{-4}{3}+\frac{1}{8}\)
=\(\frac{-29}{24}\)
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
\(3C=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\)
\(3C=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{30-27}{27.28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}+\frac{1}{28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{28.29.30}\Rightarrow C=\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right):3\)
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}:\frac{1}{\frac{7}{2}}=\frac{2}{7}:\frac{2}{7}=1\)
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)
\(\left(75\%+\frac{1}{2}\right)\text{ : }\left|-1\frac{1}{2}\right|-\frac{5}{27}\cdot1,8\cdot2012^0\)
\(=\left(\frac{3}{4}+\frac{1}{2}\right):\left|-\frac{3}{2}\right|-\frac{5}{27}\cdot\frac{9}{5}\)
\(=\frac{5}{4}\text{ : }\frac{3}{2}-\frac{1}{3}\)
\(=\frac{5}{6}-\frac{1}{3}\)
\(=\frac{1}{2}\)
\(\left(75\%+\frac{1}{2}\right)-\left|-1\frac{1}{2}\right|-\frac{5}{27}\cdot1,8\cdot2012^0\)
\(=\left(\frac{75}{100}+\frac{1}{2}\right)-\left|-\frac{3}{2}\right|-\frac{5}{27}\cdot\frac{9}{5}\cdot1\)
\(=\left(\frac{3}{4}+\frac{1}{2}\right)-\frac{3}{2}-\frac{5}{27}\cdot\frac{9}{5}\)
\(=\left(\frac{3}{4}+\frac{2}{4}\right)-\frac{3}{2}\cdot\frac{1}{3}\)
\(=\frac{5}{4}-\frac{1}{2}=\frac{5}{4}-\frac{2}{4}=\frac{3}{4}\)
Ez mà :)))