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\(\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\cdot x^2=\dfrac{8}{5}\)

=>\(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot x^2=\dfrac{8}{5}\)

=>\(\left(\dfrac{1}{2}-\dfrac{1}{10}\right)x^2=\dfrac{8}{5}\)

=>\(x^2=\dfrac{8}{5}:\dfrac{2}{5}=4\)

=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

5 tháng 5 2023

Ta có thể viết lại dãy số trong tổng bằng cách sử dụng công thức chung cho tổng của một dãy số hình học:
1/2.3 + 1/3.4 + 1/4.5 + + 1/9.10 =
(1/2 - 1/3) + (1/3 - 1/4) + ... + (1/9 -
1/10)
= 1/2 - 1/10
= 2/5
Do đó, ta có phương trình:
(1/2.3+1/3.4+1/4.5+...+1/9.10).x2 = 2/5. x2 = 8/5
Giải phương trình này ta được:
x2 = (8/5)/(2/5) = 4
Vậy, x = 2.

5 tháng 5 2023

x=2

 

30 tháng 8 2021

\(E=2.3+3.4+4.5+3.6+2.7+4.15=2\left(3+7\right)+3\left(4+6\right)+4\left(5+15\right)=2.10+3.10+4.20=20+30+80=130\)

\(F=3\left(12+13+14+15\right)+3\left(8+7+6+5\right)=3\left(12+8+13+7+14+6+15+5\right)=3\left(20+20+20+20\right)=3.80=240\)

f: Ta có: \(E=3\cdot\left(12+13+14+15\right)+3\left(8+7+6+5\right)\)

\(=3\left(12+13+14+15+8+7+6+5\right)\)

\(=3\cdot80=240\)

21 tháng 7 2023

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

15 tháng 7 2019

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)

\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)

\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)

\(\Leftrightarrow x+\frac{103}{50}=5\)

\(\Leftrightarrow x=5-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{147}{50}\)

12 tháng 8 2023

\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{2}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(x+1=20\)

\(x=20-1\)

\(x=19\)

12 tháng 8 2023

Có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(\Rightarrow x+1=20\Leftrightarrow x=19\)

12 tháng 8 2023

Lớp 5 không dùng dấu ⇒ ⇔