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giải giùm tớ nha
DH
14 tháng 5 2017
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
JH
15 tháng 5 2017
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
T
9 tháng 4 2022
Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
18 tháng 4 2017
Giải:
|
Tên tam giác |
Tên 3 đỉnh |
Tên 3 góc |
Tên 3 cạnh |
|
|
A,B,I |
|
AB, BI, IA |
|
|
A,I,C |
|
AI, IC, CA |
|
|
A,B,C |
|
AB, BC, CA |
giúp tớ với
MV
15 tháng 5 2017
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-\left(2+1+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{200-2-1-\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-...-\dfrac{2}{100}}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =2\)
15 tháng 5 2017
Đề nhỏ quá!! mà t 4 mắt. cẩn thận
Đặt :
\(A=\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+.............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....................+\dfrac{99}{100}}\)
\(A=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+..............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+.................+1-\dfrac{1}{100}}\)
\(A=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+..................+\dfrac{2}{100}\right)}{\left(1+1+.....+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+...........+\dfrac{1}{100}\right)}\)
\(A=\dfrac{2\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.............+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..............+\dfrac{1}{100}\right)}\)
\(A=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...............+\dfrac{99}{100}}=2\rightarrowđpcm\)
SG
4
18 tháng 4 2017
Giải:
|
Hình |
Tên góc (cách viết thông thường) |
Tên đỉnh |
Tên cạnh |
Tên góc (Cách viết kí hiệu) |
|
a |
Góc yCz, góc zCy, góc C |
C |
Cy,Cz |
|
|
b |
Góc MTP, PTM, T Góc TMP, PMT,M Góc TPM, MPT,P |
T M P |
TM,TP MT,MP PT,PM |
|
|
c |
Góc xPy,yPx,P Góc ySz,zSy |
P S |
Px, Py Sy, Sz |
|
\(1,x-\dfrac{2}{7}=\dfrac{4}{3};x=\dfrac{4}{3}+\dfrac{2}{7}=\dfrac{34}{21}\)
\(2,x+\dfrac{3}{4}=\dfrac{6}{8};x=\dfrac{6}{8}-\dfrac{3}{4}=0\)
\(3,\left|x\right|-\dfrac{3}{4}=\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{11}{12};\left|x\right|=\dfrac{11}{12}+\dfrac{3}{4};\left|x\right|=\dfrac{5}{3}\Rightarrow x=\left[{}\begin{matrix}\dfrac{-5}{3}\\\dfrac{5}{3}\end{matrix}\right.\)
\(4,\left(x+1\right).3=4.5=20;x+1=\dfrac{20}{3}\Leftrightarrow x=\dfrac{17}{3}\)
thanks