A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{101}{\left(50.51\right)^2}\)
= \(\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{101}{2500.2601}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{2500}-\frac{1}{2601}\)
= \(1-\frac{1}{2601}=\frac{2600}{2601}\)
CMR:\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^2}< 1\)
\(\frac{1}{2}\left(\frac{3}{2}x+\frac{5}{4}\right)-\frac{1}{3}=\left(\frac{1}{4}\right)\)
Tìm x dùm mình nha, mình sắp nộp rồi
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x
Tĩm X?
Rút gọn
C = \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
D = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
E = \(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Cho\(\frac{a}{b}=\frac{c}{d}\)
Chứng minh
\(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
Giúp mình nha mình đanh cần gấp
1.Tìm x,biết:
\(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)
2.Cho a,b,c \(\ne0\) thỏa mãn a+b-c=0.Tính \(\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
Giúp mình với,mình gần kiểm tra rồi.
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x=\frac{1}{20}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
2. Tìm x, y, z biết\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
3.Tìm x\(a,2009-\left|x-2009\right|=x\)
\(b,\left|3x+2\right|=\left|5x-3\right|\)
Tính
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
1, Tính
\(A=\left(-3+\frac{3}{4}-\frac{1}{3}\right):\left(5+\frac{2}{5}-\frac{2}{3}\right)\)
\(B=\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right).\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{11}{12}\right)\)
