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Nhi Nguyễn
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Nhi Nguyễn
4 tháng 4 2018 lúc 18:47

HELP ME ! MK cần gấp

Nguyễn Khánh Linh
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Xyz OLM
23 tháng 8 2019 lúc 14:47

1) Tính : 

a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).0\)

\(=0\)

2) Tìm x 

a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)

\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)

\(\Rightarrow x-1,01=1\)

\(\Rightarrow x=1+1,01\)

\(\Rightarrow x=2,01\)

Ngọc Ánh
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Trần Trần
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Cô nàng Thiên Bình
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ST
19 tháng 1 2018 lúc 21:06

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{998}{1000}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{499}{1000}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1000}\)

=>x+1=1000

=>x=999

anh chàng đẹp trai
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Lê Tài Bảo Châu
3 tháng 6 2019 lúc 8:28

Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

FAH_buồn
3 tháng 6 2019 lúc 8:35

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

2.

1/2004

sudy well

study well

Nguyễn Châu Mỹ Linh
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B.Thị Anh Thơ
8 tháng 1 2020 lúc 18:02

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

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trần thị khánh vy
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shitbo
6 tháng 2 2020 lúc 16:33

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

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Ngoc Han ♪
6 tháng 2 2020 lúc 18:02

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

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Fudo
6 tháng 2 2020 lúc 19:45

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)

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Nguyễn Thị Minh Huyền
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Trần Phúc
16 tháng 8 2017 lúc 20:40

Ta có:

\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)

\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)

Trần Phúc
16 tháng 8 2017 lúc 20:43

c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.

d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)

\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)

\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)