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\(M=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}+\sqrt{a}\right).\dfrac{1}{\sqrt{a}+1}\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\sqrt{a}+1}=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\sqrt{a}+1}\)
\(=\sqrt{a}+1\)
\(a=2020-2\sqrt{2019}=2019-2\sqrt{2019}+1=\left(\sqrt{2019}-1\right)^2\)
\(\Rightarrow\sqrt{a}=\sqrt{2019}-1\)
\(\Rightarrow M=\sqrt{a}+1=\sqrt{2019}-1+1=\sqrt{2019}\)
\(A=\sqrt{\left(2020-2x\right)^2}+\sqrt{\left(2019-2x\right)^2}-2\)
\(=\left|2020-2x\right|+\left|2019-2x\right|-2\)
\(=\left|2020-2x\right|+\left|2x-2019\right|-2\)
\(\ge\left|2020-2x+2x-2019\right|-2=\left|1\right|-2=-1\)
Dấu "=" xảy ra <=> ( 2020 - 2x )( 2x - 2019 ) ≥ 0 <=> 2019/2 ≤ x ≤ 1010
Vậy MinA = -1
a) A = \(\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right)\)
= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\)
= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2}\) = \(\sqrt{a}+1\)
b) a = \(2020-2.\sqrt{2019}\) = \(\left(\sqrt{2019}-1\right)^2\)
=> \(A=\sqrt{\left(\sqrt{2019}-1\right)^2}+1\) = \(\sqrt{2019}\)
nếu đó là câu giải pt thì bắt buộc phải đặt ĐKXĐ, nếu đó là câu rút gọn thì không cần nhé
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)
\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)
\(\Rightarrow x-y=1\Rightarrow P=1\)
\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)
\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)
\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)
\(A=\left(\sqrt{2019}-\sqrt{2020}\right)\left(\sqrt{2019}+\sqrt{2020}\right)\\ \rightarrow A=\left(\sqrt{2019}\right)^2-\left(\sqrt{2020}\right)^2\\ \rightarrow A=2019-2020\\ \rightarrow A=-1\)
Vậy \(A=-1\)
\(A=\left(\sqrt{2019}-\sqrt{2020}\right)\left(\sqrt{2019}+\sqrt{2020}\right)\)
\(=\left(\sqrt{2019}\right)^2-\left(\sqrt{2020}\right)^2\)
\(=\sqrt{2019^2}-\sqrt{2020^2}\)
\(=2019-2020\)
\(=-1\)
Vậy \(A=-1\)