Bài 8: Tính chất của dãy tỉ số bằng nhau

Võ Nguyễn Mai Hương
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Phạm Tuấn Đạt
16 tháng 12 2017 lúc 21:54

Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)

\(\Rightarrow\dfrac{x}{3}.\dfrac{1}{3}=\dfrac{y}{4}.\dfrac{1}{3};\dfrac{y}{3}.\dfrac{1}{4}=\dfrac{z}{5}.\dfrac{1}{4}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất của dãy tỉ số bằng nhau :

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)

\(\Rightarrow x=3.9=27\)

\(y=12.3=36\)

\(z=3.20=60\)

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Hải Đăng
16 tháng 12 2017 lúc 22:51

Theo bài ra ta có:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\right.\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{2.9-3.12+20}=\dfrac{6}{2}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\Rightarrow x=27\\\dfrac{y}{12}=3\Rightarrow y=36\\\dfrac{z}{20}=3\Rightarrow z=60\end{matrix}\right.\)

Vậy ..............................

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Hoàng Mạnh Thông
17 tháng 12 2017 lúc 18:31

Theo bài ra ta có:
⇒⎧⎪ ⎪⎨⎪ ⎪⎩x3=y4⇒x9=y12y3=z5⇒y12=z20⇒x9=y12=z20⇒{x3=y4⇒x9=y12y3=z5⇒y12=z20⇒x9=y12=z20

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

x9=y12=z20=2x−3y+z2.9−3.12+20=62=3x9=y12=z20=2x−3y+z2.9−3.12+20=62=3

⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x9=3⇒x=27y12=3⇒y=36z20=3⇒z=60⇒{x9=3⇒x=27y12=3⇒y=36z20=3⇒z=60

Vậy ..............................

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Nguễn Thị Cẩm Ly
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Trần Thu Huyền
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Shizadon
18 tháng 10 2017 lúc 20:19

\(\dfrac{x-2}{2018}=\dfrac{x-3}{2017}=\dfrac{x-4}{2016}=\dfrac{x-5}{2015}\)

\(\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=\dfrac{x-4}{2016}+\dfrac{x-5}{2015}\)

\(\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=\left(\dfrac{x-4}{2016}-1\right)+\left(\dfrac{x-5}{2015}-1\right)\)

\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=\dfrac{x-2020}{2016}+\dfrac{x-2020}{2015}\)

\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}-\dfrac{x-2020}{2015}=0\)

\(\left(x-2020\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)

\(\dfrac{1}{2018};\dfrac{1}{2017};\dfrac{1}{2016};\dfrac{1}{2015}>0\)

Nên \(x-2020=0\)

\(x=0+2020\)

\(x=2020\)

Vậy x bằng 2020

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hh Clroyalhh
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Vy
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Akai Haruma
10 tháng 3 2018 lúc 17:12

Lời giải:

Ta có \(A=\frac{1}{1.1981}+\frac{1}{2.1982}+...+\frac{1}{25.2005}\)

\(\Rightarrow 1980A=\frac{1980}{1.1981}+\frac{1980}{2.1982}+...+\frac{1980}{25.2005}\)

\(\Leftrightarrow 1980A=\frac{1981-1}{1.1981}+\frac{1982-2}{2.1982}+....+\frac{2005-25}{25.2005}\)

\(\Leftrightarrow 1980A=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+...+\frac{1}{25}-\frac{1}{2005}\)

\(1980A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+..+\frac{1}{2005}\right)\) (1)

Lại có:

\(25B=\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{1980.2005}\)

\(\Leftrightarrow 25B=\frac{26-1}{1.26}+\frac{27-2}{2.27}+...+\frac{2005-1980}{1980.2005}\)

\(\Leftrightarrow 25B=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{1980}-\frac{1}{2005}\)

\(25B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1980}\right)-\left(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{2005}\right)\)

\(25B=\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+...+\frac{1}{2005}\right)\) (2)

Từ \((1); (2)\Rightarrow 1980A=25B\Rightarrow \frac{A}{B}=\frac{25}{1980}=\frac{5}{396}\)

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Trần Thị Hảo
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Hoàng Anh Thư
8 tháng 3 2018 lúc 13:58

\(a^n-b^n=\left(a-b\right)\left(a+b\right)\)

VP: (a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2

để VP=VT thì n=2

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Ta Chia Tay Đi
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Trang
9 tháng 10 2017 lúc 18:43

a) \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\) \(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\\ \Rightarrow\dfrac{x}{2008}=1\\ \Rightarrow x=2008\)

b) \(\dfrac{7}{x}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{7}{15}\\ \Rightarrow x=15\)

c) \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)

\(\Rightarrow2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=\dfrac{15}{93}.2\)

\(\Rightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\\ \Rightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\\ \Rightarrow\dfrac{2x}{3\left(2x+3\right)}=\dfrac{10}{31}\\ \Rightarrow\dfrac{10.3\left(2x+3\right)}{31}=2x\\ \Rightarrow\dfrac{30\left(2x+3\right)}{31}=2x\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{31}:2\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{62}\\ \Rightarrow x=\dfrac{15\left(2x+3\right)}{31}\\\Rightarrow\dfrac{15\left(2x+3\right)}{x}=31\\ \Rightarrow\dfrac{30x+45}{x}=31\\ \Rightarrow30+\dfrac{45}{x}=31\\ \Rightarrow \dfrac{45}{x}=1\\ \Rightarrow x=45\)

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Nguyễn Thanh Hằng
9 tháng 10 2017 lúc 12:05

a/ \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-............-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+.......+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+.......+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\dfrac{3}{16}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}=\dfrac{13}{16}\)

\(\Leftrightarrow x=1631,5\)

Vậy ..................

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Chuột yêu Gạo
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Natsu Dragneel
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phạm thị phương thảo
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Nhã Doanh
24 tháng 2 2018 lúc 14:04

\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(=\dfrac{7^{48}.5^{29}.2^8\left(5-5.7.2^2\right)}{5^{29}.2^8.7^{48}}\)

\(=\dfrac{5-5.7.2^2}{1}\)

\(=5-140\)

\(=-135\)

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