(\(\dfrac{1}{15}\)+\(\dfrac{1}{35}\)+\(\dfrac{1}{63}\)) x a = 1
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VT
25 tháng 6 2023
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
TK
5
26 tháng 2 2022
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)
16 tháng 4 2017
\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
= 2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)) :2
= (\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)) : 2
= (\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}\) \(-\dfrac{1}{13}+\dfrac{1}{13}\)\(-\dfrac{1}{15}\)):2
= (\(\dfrac{1}{3}-\dfrac{1}{15}\)) :2
= \(\dfrac{4}{15}\): 2 = \(\dfrac{2}{15}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 3 2023
( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
\(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)
\(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)
\(x\) = 4
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
17 tháng 6 2021
Đặt biểu thực là A
2A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{21.23}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{21}-\dfrac{1}{23}=1-\dfrac{1}{23}=\dfrac{22}{23}\)
=> A = \(\dfrac{11}{23}\)
TK
3
30 tháng 10 2018
b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)
Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)
\(\Rightarrow t^2-t-20=0\)
\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)
Thay t=5 vào (1), ta có :
\(\sqrt{x^2-3x+7}=5\)
\(\Leftrightarrow x^2-3x+7=25\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Rightarrow x_1=6;x_2=-3\)
vậy...
28 tháng 11 2022
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{2}{\left(x+3\right)\left(x+5\right)}+\dfrac{2}{\left(x+5\right)\left(x+7\right)}+\dfrac{2}{\left(x+7\right)\left(x+9\right)}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+9}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{x+9-x-1}{\left(x+1\right)\left(x+9\right)}=\dfrac{2}{5}\)
=>2(x+1)(x+9)=5*8=40
=>x^2+9x+9=20
=>x^2+9x-11=0
hay \(x=\dfrac{-9\pm5\sqrt{5}}{2}\)
=>x^2+9x
`(1/15+1/35+1/63)xxa=1`
`(1/(3xx5)+1/(5xx7) +1/(7xx9)) xx a=1`
`1/9 xx a=1`
`a=1:1/9`
`a=1xx9`
`a=9`
cảm ơn bạn Lê Michael nha!