N = \(\frac{2}{5}\)+ \(\frac{2}{25}\)+ \(\frac{2}{125}\)+ \(\frac{2}{625}\)+ \(\frac{2}{3125}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
= \(1+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
= \(1+\frac{3}{4}\)
= \(\frac{4}{4}+\frac{3}{4}\)
= \(\frac{7}{4}\)
HỌC TỐT
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(\frac{17}{6}+\frac{13}{9}\right):\left(\frac{121}{12}+\frac{19}{2}\right)\)
\(=\frac{77}{18}:\frac{235}{12}=\frac{77}{18}.\frac{12}{235}=\frac{154}{705}\)
b, \(\left(\frac{19}{2}-\frac{35}{4}\right).\frac{7}{2}+\frac{5}{8}:\frac{5}{3}\)
\(\frac{3}{4}.\frac{7}{2}+\frac{5}{8}.\frac{3}{5}=\frac{21}{8}+\frac{3}{8}=\frac{24}{8}=3\)
c, \(140.\left(\frac{25}{28}+\frac{25}{36}+\frac{5}{9}\right)=140.\frac{10}{7}=200\)
d, \(\left(\frac{7}{5}-\frac{7}{25}+\frac{7}{125}-\frac{7}{625}\right):\left(1-\frac{1}{5}+\frac{1}{25}-\frac{1}{125}\right)\)
\(=\frac{728}{625}:\frac{104}{125}=\frac{728}{625}.\frac{125}{104}=\frac{7}{5}\)
c)=140[(50.10101)/(56.10101)+(50.10101)/(72.10101)+(50.10101)/(90.10101)]
=140(50/56+50/72+50/90)
=140.50(1/56+1/72+1/90)
=7000(1/7.8+1/8.9+1/9.10)
=7000(1/7-1/8+1/8-1/9+1/9-1/10)
=7000(1/7-1/10)
=7000.3/70
=300
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tham khảo: Câu hỏi của Lê Thị Ngọc Duyên - Toán lớp 9 | Học trực tuyến
\(N=\frac{2}{5}+\frac{2}{25}+\frac{2}{125}+\frac{2}{625}+\frac{2}{3125}\)
\(5\times N=2+\frac{2}{5}+\frac{2}{25}+\frac{2}{125}+\frac{2}{625}\)
\(5\times N-N=\left(2+\frac{2}{5}+\frac{2}{25}+\frac{2}{125}+\frac{2}{625}\right)-\left(\frac{2}{5}+\frac{2}{25}+\frac{2}{125}+\frac{2}{625}+\frac{2}{3125}\right)\)
\(4\times N=2-\frac{2}{3125}\)
\(N=\frac{1562}{3125}\)