1) Tìm nghiệm nguyên của phương trình : xy+y = x3 +x2 +7
2) Giải phương trình : \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{2.2011}{2012}\)
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22 tháng 3 2021
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
12 tháng 11 2023
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
MH
30 tháng 5 2022
Thay \(x=\dfrac{3}{4}y\) vào phương trình dưới, ta có:
\(\dfrac{1}{2}\left(\dfrac{3}{4}y+3\right)\left(y-2\right)-\dfrac{1}{2}.\dfrac{3}{4}y^2=9\)
\(\Leftrightarrow\dfrac{3}{8}y^2-\dfrac{3}{4}y+\dfrac{3}{2}y-3-\dfrac{3}{8}y^2=9\\ \Leftrightarrow\dfrac{3}{4}y=12\\ \Leftrightarrow y=18\Rightarrow x=12\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(12;18\right)\)
Y
9 tháng 4 2022
+ Pt thứ nhất :
Ta có mẫu thức chung là : \(2\left(x-3\right)\left(x+1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x\ne2\\x-3\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne3\\x\ne-1\end{matrix}\right.\)
Vậy \(ĐKXĐ\) là :\(x\ne2;3;-1\)
+ Pt thứ hai :
Ta có mẫu thức chung là : \(\left(x-2\right)\left(x+3\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)
Vậy \(DKXD:\) \(\) \(x\ne2;-3\)
MH
17 tháng 10 2023
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
17 tháng 10 2023
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
1/
\(y\left(x+1\right)-x^2\left(x+1\right)=7\Leftrightarrow\left(x+1\right)\left(y-x^2\right)=7\)
TH1: \(\left\{{}\begin{matrix}x+1=1\\y-x^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=7\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+1=7\\y-x^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=37\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x+1=-1\\y-x^2=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x+1=-7\\y-x^2=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=63\end{matrix}\right.\)
2/
\(\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)...\left(1+\dfrac{1}{\left(x+1-1\right)\left(x+1+1\right)}\right)=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{\left(x+1\right)^2}{x\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{2.3.4...\left(x+1\right)}{3.4.5...\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow2012\left(x+1\right)=2011\left(x+2\right)\)
\(\Leftrightarrow x=2010\)