\(\dfrac{-18}{42}\)x\(\dfrac{35}{42}\):\(\dfrac{-98}{42}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
14 tháng 2 2022
\(\dfrac{7}{21}+\dfrac{-9}{36}=\dfrac{1}{3}+\dfrac{-1}{4}=\dfrac{4}{12}+\dfrac{-3}{12}=\dfrac{1}{12}\)
\(\dfrac{-12}{18}+\dfrac{-21}{35}=\dfrac{-2}{3}+\dfrac{-3}{5}=\dfrac{-10}{15}+\dfrac{-9}{15}=\dfrac{-19}{15}\)
\(\dfrac{-18}{14}+\dfrac{15}{-21}=\dfrac{-9}{7}+\dfrac{-5}{7}=\dfrac{-14}{7}=-2\)
\(\dfrac{3}{21}+\dfrac{-6}{42}=\dfrac{1}{7}+\dfrac{-1}{7}=0\)
S
3
27 tháng 2 2022
\(=\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{2}-\dfrac{5}{6}+2022=2022\)
PV
1
22 tháng 3 2021
$ĐKXĐ:x \neq -4;-5;-6;-7$
$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$
$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$
$⇔x^2+11x+28=54$
$⇔x^2+11x-26=0$
$⇔x^2-2x+13x-26=0$
$⇔(x-2)(x+13)=0$
$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)
Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$
BC
1
30 tháng 4 2021
Ta có:
\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)
pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\) (tm)
b)
29 tháng 10 2023
a: 1/2 của 18 là \(\dfrac{1}{2}\text{x}18=\dfrac{18}{2}=9\)
1/7 của42 là \(\dfrac{1}{7}\text{x}42=\dfrac{42}{7}=6\)
b: 3/4 của 20 là \(\dfrac{3}{4}\text{x}20=\dfrac{3\text{x}20}{4}=\dfrac{60}{4}=15\)
3/7 của 21 là \(\dfrac{3}{7}\text{x}21=\dfrac{3\text{x}21}{7}=\dfrac{63}{7}=9\)
c:2/5 của 35 là: \(35\text{x}\dfrac{2}{5}=\dfrac{35\text{x}2}{5}=14\)
2/9 của 36 là \(\dfrac{2}{9}\text{x}36=\dfrac{2\text{x}36}{9}=\dfrac{72}{9}=8\)
NG
Người Già
CTVVIP
24 tháng 8 2023
a) \(\dfrac{36}{42}\) = \(\dfrac{18}{21}\) = \(\dfrac{6}{7}\) = \(\dfrac{30}{35}\)
b) \(\dfrac{80}{100}\) =\(\dfrac{16}{20}\) = \(\dfrac{4}{5}\) = \(\dfrac{40}{50}\)
13 tháng 9 2023
d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7
\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>x^2+11x+28=54
=>x^2+11x-26=0
=>(x+13)(x-2)=0
=>x=2 hoặc x=-13
e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)
=>x-258=0
=>x=258
PA
8 tháng 3 2017
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) (ĐKXĐ: \(x\notin\left\{-4;-5;-6;-7\right\}\))
<=> \(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}-\dfrac{1}{18}=0\)
<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)
<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)
<=> \(\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}=0\)
=> \(18\left(x+7\right)-18\left(x+4\right)-\left(x+4\right)\left(x+7\right)=0\)
<=> 18x + 18.7 - 18x - 18.4 - x2 - 7x - 4x - 28 = 0
<=> - x2 - 11x + 26 = 0
<=> (x - 2)(x + 13) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\) (nhận)
Vậy S = {-13; 2}
\(-\dfrac{18}{42}\cdot\dfrac{35}{42}:\dfrac{-98}{42}\)
\(=-\dfrac{18}{42}\cdot\dfrac{35}{42}\cdot\dfrac{42}{-98}\)
\(=\dfrac{-18}{42}\cdot\dfrac{-5}{14}\\ =\dfrac{15}{98}\)
\(\dfrac{-18}{42}.\dfrac{3\text{5}}{42}:\dfrac{-98}{42}\)
\(=\dfrac{-18}{42}.\dfrac{3\text{5}}{42}.\dfrac{-42}{98}\)
\(=\dfrac{1\text{5}}{98}\)