Tìm hai số x và y biết
Trình bày chi tiết và giải theo tính chất dãy tỉ số bằng nhau nha
a) x phần y =5
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Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)
=> x + 2022 = 0
=> x = -2022
Vậy x = -2022
\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\)
\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-2022\)
\(x=-2022\)
a) \(A=0,5-\left|x-3,5\right|\le0,5\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x-3,5\right|=0\Rightarrow x=3,5\)
Vậy Max(A) = 0,5 khi x = 3,5
b) \(C=1,7+\left|3,4-x\right|\ge1,7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|3,4-x\right|=0\Rightarrow x=3,4\)
Vậy Min(C) = 1,7 khi x = 3,4
Ta có: \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)
\(\Leftrightarrow\frac{2x-2}{3}=7x+3\)
\(\Leftrightarrow2x-2=21x+9\)
\(\Leftrightarrow19x=-11\)
\(\Rightarrow x=-\frac{11}{19}\)
Ta có: \(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)
\(\Leftrightarrow\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)
Mà \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)
\(\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)
\(\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+1=0\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-1=-1\)
Ta có: \(\left(\frac{2}{3}-x\right)^2=\left(-2x+\frac{1}{3}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}-x=-2x+\frac{1}{3}\\\frac{2}{3}-x=2x-\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
\(\frac{x-1}{x-2}=\frac{x-2}{x+3}\)ĐK : \(x\ne-3;2\)
\(\Leftrightarrow\left(x-2\right)^2=\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x^2-4x+4=x^2-3x-x+4\)
\(\Leftrightarrow0=0\)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
<=> ( x - 1 ) ( x + 3 ) = ( x - 2 ) ( x + 2 )
<=> x2 + 2x - 3 = x2 - 4
<=> x2 + 2x - 3 - x2 + 4 = 0
<=> 2x + 1 = 0
<=> 2x = - 1
<=> x = - 1/2