Tìm GTNN:
B = \(\sqrt{\left(x-2020\right)^2}+\sqrt{\left(x-1^2\right)}\)
K
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S
29 tháng 12 2020
\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P\left(x\right)=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(P\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(P\left(x\right)=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2\)
\(P\left(x\right)=x-\sqrt{x}\)
Ta có : \(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{x-\sqrt{x}}{2020\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2020\sqrt{x}}=\dfrac{\sqrt{x}-1}{2020}\)
Để \(\dfrac{P\left(x\right)}{2020\sqrt{x}}min\Leftrightarrow\dfrac{\sqrt{x}-1}{2020}min\Leftrightarrow\sqrt{x}-1\) min (vì 2020 > 0)
Lại có : \(\sqrt{x}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> x = 0
Vậy Min\(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{-1}{2020}\Leftrightarrow x=0\)
27 tháng 6 2021
`(x+sqrt{x^2+2020})(sqrt{x^2+2020}-x)=x^2+2020-x^2=2020`
`=>y+sqrt{y^2+2020}=sqrt{x^2+2020}-x`
`<=>x+y=sqrt{x^2+2020}-sqrt{y^2+2020}`
Tương tự:`x+y=sqrt{y^2+2020}-sqrt{x^2+2020}`
Cộng từng vế
`=>2(x+y)=0`
`<=>S=0+2020=2020`
27 tháng 6 2021
Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(x-\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2020\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+2020}=x-\sqrt{x^2+2020}\) (1)
Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(y-\sqrt{y^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)
\(\Leftrightarrow\left(y^2-y^2-2020\right)\left(x+\sqrt{x^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)
\(\Leftrightarrow-x-\sqrt{x^2+2020}=y-\sqrt{y^2+2020}\) (2)
Từ (1) (2) cộng vế với vế \(\Rightarrow-\left(x+y\right)-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)=x+y-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
\(S=x+y+2020=2020\)
12 tháng 7 2021
Bài 1.
Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)
\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)
\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\) (1)
Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)
\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)
\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\) (2)
Cộng vế với vế của (1) và (2) ta có:
\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)
\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
12 tháng 7 2021
Bài 2:
Ta có: (2a+1)(2b+1)=9
nên \(2b+1=\dfrac{9}{2a+1}\)
\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)
\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)
\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)
Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)
\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)
\(=\dfrac{3+2a+1}{3a+6}\)
\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{\left(2y\right)^2+2020}\right)=2020\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+\sqrt{\left(2y\right)^2+2020}=\sqrt{x^2+2020}-x\\x+\sqrt{x^2+2020}=\sqrt{\left(2y\right)^2+2020}-2y\end{matrix}\right.\)
\(\Rightarrow x+2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}=-x-2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}\)
\(\Leftrightarrow2\left(x+2y\right)=0\)
\(\Leftrightarrow x=-2y\)
\(\Rightarrow B=2y^2-8y^2+3y^2-2y+3y+15\)
\(\Rightarrow B=-3y^2+y+15=-3\left(y-\dfrac{1}{6}\right)^2+\dfrac{181}{12}\)
\(B_{max}=\dfrac{181}{12}\) khi \(y=\dfrac{1}{6}\)
23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
BH
27 tháng 7 2019
\(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2020}{x+\sqrt{x^2+2020}}=y+\sqrt{y^2+2020}\\\frac{2020}{y+\sqrt{y^2+2020}}=x+\sqrt{x^2+2020}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+\sqrt{x^2+2020}=y+\sqrt{y^2+2020}\\-y+\sqrt{y^2+2020}=x+\sqrt{x^2+2020}\end{cases}}\)
\(\Leftrightarrow-2x-2y=0\)(cộng 2 vế )
\(\Leftrightarrow x+y=0\)
B
27 tháng 7 2019
Mềnh còn cách khác:)
\(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Ta có:\(\left(\sqrt{x^2+2020}+x\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)
Lại có:\(\left(\sqrt{x^2+2020}+x\right)\left(\sqrt{y^2+2020}+y\right)=2020\)
\(\Rightarrow\sqrt{x^2+2020}-x=\sqrt{y^2+2020}+y\)
\(\Leftrightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)(1)
\(\left(\sqrt{y^2+2020}+y\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)
\(\Rightarrow\sqrt{y^2+2020}-y=\sqrt{x^2+2020}+x\)
\(\Leftrightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)(2)
Cộng vế với vế của (1) và (2) ta có:\(x+y+x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}+\sqrt{y^2+2020}-\sqrt{x^2+2020}\)
\(\Leftrightarrow2x+2y=0\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
27 tháng 7 2019
\(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(x-\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2020\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-2020\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+2020}=x-\sqrt{x^2+2020}\)
\(\Leftrightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)(1)
Chứng minh tương tự ta cũng có \(x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)(2)
Cộng theo vế của (1) và (2) ta được :
\(2\left(x+y\right)=\sqrt{x^2+2020}-\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{y^2+2020}\)
\(\Leftrightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
Vậy...
VD
1
VH
24 tháng 9 2020
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Câu hỏi của Chuột yêu Gạo - Toán lớp 9 | Học trực tuyến
\(B=\sqrt{\left(x-2020\right)^2}+\sqrt{\left(x-1\right)^2}=\left|x-2020\right|+\left|x-1\right|\)
\(=\left|x-2020\right|+\left|1-x\right|\ge\left|x-2020+1-x\right|=\left|-2019\right|=2019\)
Dấu " = " xảy ra \(\Leftrightarrow\left(x-2020\right)\left(1-x\right)\ge0\)
TH1: \(\hept{\begin{cases}x-2020\le0\\1-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2020\\1\le x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2020\\x\ge1\end{cases}}\Leftrightarrow1\le x\le2020\)
TH2: \(\hept{\begin{cases}x-2020>0\\1-x>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>2020\\1>x\end{cases}}\Leftrightarrow\hept{\begin{cases}x>2020\\x< 1\end{cases}}\)( không thỏa mãn )
Vậy \(minB=2019\)\(\Leftrightarrow1\le x\le2020\)
\(B=|x-2020|+|x-1|\)
\(=|2020-x|+|x-1|\)
Áp dụng bất đẳng thức giá trị tuyệt đối :
\(|2020-x|+|x-1|\ge|2020-x+x-1|\)
\(|2020-x|+|x-1|\ge|2020-x+x-1|\)
\(|2020-x|+|x-1|\ge|2019|\)
\(|2020-x|+|x-1|\ge2019\)
Dấu = xảy ra \(\Leftrightarrow\) \(\left(2020-x\right)\left(x-1\right)\ge0\)
Có 2 TH
TH 1 :
\(\hept{\begin{cases}2020-x\ge0\\x-1\ge0\end{cases}}\)
\(\hept{\begin{cases}-x\ge-2020\\x\ge1\end{cases}}\)
\(\hept{\begin{cases}x\le2020\\x\ge1\end{cases}\Rightarrow1\le x\le2020}\)
TH2 :
\(\hept{\begin{cases}2020-x\le0\\x-1\ge0\end{cases}}\)
\(\hept{\begin{cases}-x\le-2020\\x\le1\end{cases}}\)
\(\hept{\begin{cases}x\ge2020\\x\le1\end{cases}\Rightarrow x=\varnothing}\)