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\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)
\(=\frac{6}{1x4}+\frac{6}{4x7}+\frac{6}{7x10}+\frac{6}{10x13}\)
\(=2\left(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+\frac{3}{10x13}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(=2\left(1-\frac{1}{13}\right)\)
\(=2x\frac{12}{13}\)
\(=\frac{24}{13}\)
\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}\)
\(=2\left(\frac{3}{1.4}+\frac{3}{1.7}+\frac{3}{7.10}+\frac{3}{10.13}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(\Leftrightarrow2=\left(1-\frac{1}{13}\right)\)
\(=2.\frac{12}{13}\)
\(=\frac{24}{13}\)
A=\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
A=\(\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+\frac{6}{13\cdot16}\)
A:2=\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
A:2=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\)\(\frac{1}{16}\)
A:2=\(1-\frac{1}{16}\)
A:2=\(\frac{15}{16}\)
A=\(\frac{15}{8}\)
vậy ...
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{101}+\frac{1}{102}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{51}\)
\(=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
\(=VP\)
a) ta có \(\frac{-5}{6}\)\(\times\)\(\frac{120}{25}\)< \(x\)<\(\frac{-7}{15}\)\(\times\)\(\frac{4}{9}\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2074074074\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)( -1;-2;-3)
b) ta có \(\left(\frac{-5}{3}\right)^3\)<\(x\)<\(\frac{-25}{35}\)\(\times\)\(\frac{-5}{6}\)\(\Rightarrow\)\(-4,62962963\)<\(x\)<\(0,5952380952\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)(-4;-3;-2;-1;0)
ĐÚNG THÌ K CHO MK NHA
1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10
=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)
=9/10x(1/3+3/5+5/7+7/9)
9/10x(1/3+3/5)+(5/7+7/9)
=9/10x1/5+5/9
9/50+5/9
=10
Bn Long làm đúng rồi bn 
nguyễn kim arica cứ làm theo cách đó là được .
Bn nào thấy đúng thì ủng hộ nha .
2/7 + 7/-5 + -2/32 = -41/35
4/3 + | 5\-6 | + | -7/-6 | = 4/3 + 5/6 + 7/6 = 10/3
hok tốt
.............
\(\frac{2}{7}+\frac{7}{-5}+\frac{-2}{35}\)
\(=\frac{2}{7}+\frac{-7}{5}+\frac{-2}{35}\)
\(=\frac{10}{35}+\frac{-49}{35}+\frac{-2}{35}\)
\(=\frac{-41}{35}\)
\(\frac{4}{3}+\left|\frac{5}{6}\right|+\left|\frac{-7}{-6}\right|\)
\(=\frac{4}{3}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{8}{6}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{20}{6}=\frac{10}{3}\)
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
Bạn ơi, có chắc là 6/280 ở cuối không.
Trả lời nhanh để mink giải câu này cho
lộn \(\frac{6}{208}\) chứ ko phải 6/ 280