TÌM X
\(\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+...+\frac{1}{X.\left(X+3\right)}=\frac{1}{8}\)
So sánh A và B biết
a,A=6\(\left(x+\frac{1}{3}\right)^2\)
,B=-8-(3,75-x)2
b,A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
B=\(\left(\frac{1}{2}\right)^4\)
Bài 2: a.Tính
\(\frac{1}{4}-\frac{1}{7}=\frac{3}{4.7};_{ }_{ }\frac{1}{7}-\frac{1}{10}=\frac{3}{7.10};_{ }_{ }\frac{1}{10}-\frac{1}{13}=\frac{3}{10.13};_{ }_{ }\frac{1}{13}-\frac{1}{16}=\frac{3}{13.16};_{ }....._{ }\frac{1}{x}-\frac{1}{x+3}=\frac{3}{x\left(x+3\right)}\)Qui luật: \(\frac{m}{a\left(a+m\right)}=\frac{1}{a}-\frac{1}{a+m}\)
b. A = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{x\left(x+3\right)}\)
3A = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{x\left(x+3\right)}\)
3A = \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+3}\)
3 A = \(\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{63}{764}.3=\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{189}{764}=\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{1}{382}=\frac{1}{x+3}_{ }_{ }_{ }=>x=382-3=379\)
Bài toán gì mà có cả câu trả lời thế này ????????
Đăng lên mà trả lời luôn thế này thì đăng lên làm gì cho nó mệt
Tính D=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
E=\(\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{254.257}\)
F=\(\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}+...+\frac{5}{254.257}\)
M=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
các bn tính cho mik nha nhớ ghi cả lời giải
Tìm x:
\(\frac{\left(27\frac{5}{19}-26\frac{4}{13}\right)\left(\frac{3}{4}+\frac{15}{59}-\frac{3}{118}\right)}{\left(\frac{3}{4}+x\right)\frac{27}{33}}=\frac{\frac{1}{13.16}+\frac{1}{14.17}}{\frac{1}{13.15}+\frac{1}{14.16}+\frac{1}{15.17}}\)
Tìm x, biết:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được
Ta có:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
Tìm x, biết:
\(\frac{1}{1.4}\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)
\(1-\frac{1}{x+3}=\frac{375}{376}\)
\(\frac{x+2}{x+3}=\frac{375}{376}\)
=> x + 2 = 375
=> x = 375 - 2
=> x = 373
Bài 1 : tính hợp lý :
a) A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
b) B = \(\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)=\frac{1}{3}.\frac{18}{19}=\frac{6}{19}\) VẬY: \(A=\frac{6}{19}\)
\(B=\frac{1}{4.2.4}+\frac{1}{4.4.6}+...+\)\(\frac{1}{4.10.12}\)
\(B=\frac{1}{4}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{10.12}\right)\)\(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{10}-\frac{1}{12}\right)\)
\(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{120}\right)=\frac{1}{8}.\frac{59}{120}=\frac{59}{960}\)
Bài 1 : tính hợp lý :
a) A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
b) B = \(\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
đây là toán lớp 5 cơ mà
a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)
A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)
A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))
A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)
A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))
A=\(\frac{1}{3}\)x\(\frac{18}{19}\)
A=\(\frac{6}{19}\)
mik ko hiểu cách làm của bạn cho lắm ,có thể làm rõ hơn đc ko ???
Tìm x
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
...............
đặt VT là A ta có:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{6}{19}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{6}{19}\)
\(\Rightarrow\frac{1}{3}\cdot\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{6}{19}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}\cdot3=\frac{2}{19}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{2}{19}=\frac{17}{19}\)
=>17*(x+3)=19
x+3=19/17
x=19/17-3
x=-32/17
nếu thấy đúng thì k cho mh nhaThái Lâm Hoàng