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#)Giải :
\(\left(x+\sqrt{x^2+2019}\right)\left(x+\sqrt{y^2+2019}\right)=2019\)
\(\Leftrightarrow x^2+2019-x^2=2019\)
\(\Leftrightarrow\sqrt{x^2+2019}-x=\sqrt{y^2+2019}+y\)
\(\Leftrightarrow x+y=\sqrt{x^2+2019}-\sqrt{y^2+2019}\left(1\right)\)
\(\left(\sqrt{x^2+2019}+y\right)\left(\sqrt{y^2+2019}-y\right)=2019\)
\(\Leftrightarrow\sqrt{y^2+2019}-y=\sqrt{x^2+2019}+x\)
\(\Leftrightarrow x+y=\sqrt{y^2+2019}-\sqrt{x^2+2019}\left(2\right)\)
Cộng hai vế (1) và (2) với nhau. ta được :
\(2\left(x+y\right)=0\Leftrightarrow x+y=0\)
|*Đúng k nhỉ ???*|
Bài 1: Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{1}{\sqrt{2020}+\sqrt{2019}};\)\(\sqrt{2018}-\sqrt{2017}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2020}+\sqrt{2019}>\sqrt{2018}+\sqrt{2017}\)nên \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Suy ra\(\sqrt{2020}-\sqrt{2019}< \sqrt{2018}-\sqrt{2017}\)
Bài 2: Xét biểu thức \(\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}=\sqrt{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)(Vì \(a^2+a+1>0\forall a\inℝ\))
Áp dụng công thức tổng quát trên, ta được: \(\sqrt{2019^2+2019^2.2020^2+2020^2}=2019^2+2019+1\)(là số tự nhiên) (đpcm)
\(^6\sqrt{2019} = b, ^6\sqrt{2020} = a \\ Then, A = a^3 - b^3; B = a^2 -b^2\\ \Rightarrow A > B \)
Ta có: \(a^4+b^4+c^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{\left(\frac{\left(a+b+c\right)^2}{3}\right)^2}{3}=3\)
=> \(3abc\ge3\)=> \(abc\ge1\) ( 1)
Lại có: \(a^4+b^4+c^4+1\ge4\sqrt[4]{a^4b^4c^4}=4\left|abc\right|=4abc\)
=> \(3abc+1\ge4abc\Rightarrow abc\le1\)(2)
Từ (1); (2) => abc = 1
khi đó a = b = c = 1
=> P = 1^2019 + 1 ^2019 + 1^2019 = 3
\(DK:x\ge1\)
\(A=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}+2019\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|+2019\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|+2019\ge|\sqrt{x-1}+1+1-\sqrt{x-1}|+2019=2021\)
Dau '=' xay ra khi \(\left(\sqrt{x-1}+1\right)\left(1-\sqrt{x-1}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x-1}+1\ge0\\1-\sqrt{x-1}\ge0\end{cases}\Leftrightarrow x=2\left(n\right)}\)
TH2:
\(\hept{\begin{cases}\sqrt{x-1}+1\le0\\1-\sqrt{x-1}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}\le-1\\\sqrt{x-1}\ge1\end{cases}\left(l\right)}}\)
Vay \(A_{min}=2021\)khi \(x=2\)
\(\sqrt{9\left(b-2\right)^2}=\sqrt{9}.\sqrt{\left(b-2\right)^2}=3.\left|b-2\right|\)
\(\sqrt{a^2\left(a+1\right)^2}=\sqrt{a^2}.\sqrt{\left(a+1\right)^2}=\left|a\right|.\left|a+1\right|\) Nhưng do a > 0
Nên: \(\left|a\right|.\left|a+1\right|=a.\left(a+1\right)=a^2+a\)
\(\sqrt{b^2\left(b-1\right)^2}=\sqrt{b^2}.\sqrt{\left(b-1\right)^2}=\left|b\right|.\left|\left(b-1\right)\right|\)
Em mới lớp 5 thôi sai đừng trách :v
Chúc anh học tốt !!!
Nhân liên hợp là ra -.-
a, Có: \(\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{a^2+2019}-a\right)=a^2+2019-a^2=2019\)
Mà \(\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{b^2+2019}+b\right)=2019\)
\(\Rightarrow\sqrt{a^2+2019}-a=\sqrt{b^2+2019}+b\)(1)
b,Tương tự câu a sẽ c/m được \(\sqrt{a^2+2019}+a=\sqrt{b^2+2019}-b\)(2)
Lấy (1) trừ (2) theo từng vế được
\(\sqrt{a^2+2019}-a-\sqrt{a^2+2019}-a=\sqrt{b^2+2019}+b-\sqrt{b^2+2019}+b\) \(\Leftrightarrow-2a=2b\)
\(\Leftrightarrow-a=b\)
\(\Rightarrow-a^{2019}=b^{2019}\)
Ta có: \(P=a^{2019}+b^{2019}+2019\)
\(=a^{2019}-a^{2019}+2019\)
\(=2019\)
a)Theo giả thiết thì \(VT=\frac{\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{a^2+2019}-a\right)}{\sqrt{a^2+2019}-a}.\frac{\left(\sqrt{b^2+2019}+b\right)\left(\sqrt{b^2+2019}-b\right)}{\sqrt{b^2+2019}-b}=2019\)
\(\Leftrightarrow\frac{2019}{\sqrt{a^2+2019}-a}.\frac{2019}{\sqrt{b^2+2019}-b}=2019\)
\(\Rightarrow\frac{1}{\sqrt{a^2+2019}-a}.\frac{1}{\sqrt{b^2+2019}-b}=1\) (chia hai vế cho 2019)
Suy ra \(\sqrt{a^2+2019}-a=\sqrt{b^2+2019}-b\)?!? (lạ nhỉ,hay là tui làm sai gì đó chăng?)