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Câu 1 :
\(x:\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{101.103}\right)=1\)
\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\right]\) \(=1\)
\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{103}\right)\right]=1\)
\(=>\) \(x:\frac{51}{103}=1\)
\(=>x=1.\frac{51}{103}=\frac{51}{103}\)
Câu 2 :
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\right).x=2\)
\(=>\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right).x=2\)
\(=>\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)
\(=>\frac{11}{12}.x=2\)
\(=>x=2:\frac{11}{12}\)
\(=>x=\frac{24}{11}\)
\(A=\frac{1^2}{1\times2}\times\frac{2^2}{2\times3}\times\frac{3^2}{3\times4}\times\frac{4^2}{4\times5}\)
\(=\frac{1}{2}\times\frac{4}{6}\times\frac{9}{12}\times\frac{16}{20}\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)
Gạch các số giống nhau của phép nhân đó là 2; 3; 4. Ta được kết quả bằng
\(=\frac{1}{5}\)
1-1/x+1=2015/2016
=>1/x+1=1-2015/2016=1/2016
=>x+1=2016=>x=2015
mình không ghi lại đề nha:
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
<=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
<=>\(\frac{x}{x+1}=\frac{2015}{2016}\)
=>x=
Đến đó bạn tự giải tiếp ha
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)
B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)
= \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)
= \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)
= \(\frac{-22}{105}\)
C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
= 1/2 .( 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + .......+ 1/2014.2015 - 1/2015.2016)
= 1/2 ( 1/2 - 1/2015.2016)
Tính tiếp p nhé.
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
= 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/x - 1/x + 1
= 1/2 - 1 /x + 1 =199/200
=1/x+1 = 1/2 - 199/200
1/ x + 1 = ....
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=-\frac{99}{200}\)
\(\Rightarrow\frac{-99}{-\left(x+1\right).99}=\frac{-99}{200}\)
\(\Rightarrow-99x+-99=200\)
a, \(\frac{4x}{3}=\frac{14x}{3}+5\)
\(\frac{4x}{3}-\frac{14x}{3}=5\)
\(\frac{-10x}{3}=5\)
x=-1,5
b, 3x-5x=2-4
-2x=-2
x=1
c, \(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\) .x=2
\(\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)
\(\frac{11}{12}.x=2\)
x=\(\frac{24}{11}\)
d, \(x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)
\(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
x=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)
x=\(\frac{1}{2}.\frac{102}{103}\)
x=\(\frac{51}{103}\)