\(\dfrac{300}{x}\) +1=\(\dfrac{400}{x+1}\)
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14 tháng 3 2017
\(\left(\dfrac{1}{200}+\dfrac{1}{300}+\dfrac{1}{400}+...+\dfrac{1}{1000}\right).1.2.3.4.5\).\(\left(\dfrac{1}{120}-\dfrac{1}{180}-\dfrac{1}{360}\right)\)
Xét \(\left(\dfrac{1}{120}-\dfrac{1}{180}-\dfrac{1}{360}\right)\) ta có :
= \(\dfrac{1}{120}-\left(\dfrac{1}{180}+\dfrac{1}{360}\right)\) =\(\dfrac{1}{120}-\dfrac{1}{120}=0\)
Trong 1 tích nếu có 1 thừa số 0 thì tích đó bằng 0
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1 tháng 2 2021
Ta có: \(\left\{{}\begin{matrix}x-y=10\\\dfrac{300}{y}-\dfrac{300}{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\dfrac{300}{y}-\dfrac{300}{10+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\dfrac{300\left(y+10\right)}{y\left(y+10\right)}-\dfrac{300y}{y\left(y+10\right)}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\300y+3000-300y=y\left(y+10\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\y^2+10y-3000=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\y^2+10y+25-3025=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\left(y+5\right)^2=3025\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y+10\\x=y+10\end{matrix}\right.\\\left[{}\begin{matrix}y+5=55\\y+5=-55\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=50+10=60\\x=-60+10=-50\end{matrix}\right.\\\left[{}\begin{matrix}y=50\\y=-60\end{matrix}\right.\end{matrix}\right.\)
Vậy: Hệ phương trình có hai cặp nghiệm là (x,y)\(\in\){(-50;-60);(60;50)}
LN
3
HH
4 tháng 2 2018
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
LT
3
IM
21 tháng 4 2017
Ta có :
\(\dfrac{1}{3^{400}}=\dfrac{1}{\left(3^4\right)^{100}}=\dfrac{1}{81^{100}}\)
\(\dfrac{1}{4^{300}}=\dfrac{1}{\left(4^3\right)^{100}}=\dfrac{1}{64^{100}}\)
Vì \(81^{100}>64^{100}\)
\(\Rightarrow\dfrac{1}{81^{100}}< \dfrac{1}{61^{100}}\)
Vậy ...
21 tháng 4 2017
phân số cùng tử số mà phân số nó to hơn thì phân số nó bé hơn
=>1/3^400>1/4^300
1 tháng 3 2021
`d,(10x+3)/12=1+(6+8x)/9`
`<=>(10x+3)/12=(8x+15)/9`
`<=>30x+9=32x+60`
`<=>2x=-51`
`<=>x=-51/2`
23 tháng 6 2017
Ta có :
\(A=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)...............\left(1-\dfrac{1}{361}\right)\left(1-\dfrac{1}{400}\right)\)
\(\Rightarrow A=\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\left(\dfrac{25}{25}-\dfrac{1}{25}\right).............\left(\dfrac{361}{361}-\dfrac{1}{361}\right)\left(\dfrac{400}{400}-\dfrac{1}{400}\right)\)\(\Rightarrow A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}................\dfrac{360}{361}.\dfrac{399}{400}\)
\(\Rightarrow A=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...............\dfrac{18.20}{19^2}.\dfrac{19.21}{20^2}\)
\(\Rightarrow A=\dfrac{\left(2.3.4......19\right)\left(4.5.6......21\right)}{\left(3.4.5.....20\right)\left(3.4.5...20\right)}=\dfrac{2.21}{3.20}=\dfrac{7}{10}\)
=>\(\dfrac{400}{x+1}-\dfrac{300}{x}=1\)
=>\(\dfrac{400x-300x-300}{x\left(x+1\right)}=1\)
=>x(x+1)=100x-300
=>x^2+x-100x+300=0
=>x^2-99x+300=0
=>\(\left[{}\begin{matrix}x\simeq95,87\\x\simeq3,13\end{matrix}\right.\)