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a, P>0
Có \(P^2=x+2\sqrt{x\left(2-x\right)}+2-x=2+2\sqrt{2x-x^2}=\sqrt{1-\left(x^2-2x+1\right)}+2=2+\sqrt{1-\left(x-1\right)^2}\)
Luôn có: \(1-\left(x-1\right)^2\le1\)=> \(0\le\sqrt{1-\left(x-1\right)^2}\le1\)<=> \(0\le2\sqrt{1-\left(x-1\right)^2}\le4\)
<=> \(2\le2+2\sqrt{1-\left(x-1\right)^2}\le2+2\)
<=> \(2\le P^2\le4\)
<=> \(\sqrt{2}\le P\le2\)(do P>0)
minP xảy ra <=> \(\sqrt{1-\left(x-1\right)^2}=0\)
<=> \(\left(x-1\right)^2=1\) <=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)(t/m)
maxP xảy ra<=> \(\sqrt{1-\left(x-1\right)^2}=1\)
<=> \(\left(x-1\right)^2=0\) <=> x=1(t/m)
b, Q>0 (đk :\(2019\le x\le2020\))
Có \(Q^2=x-2019+2\sqrt{\left(x-2019\right)\left(2020-x\right)}+2020-x=1+2\sqrt{\left(x-2019\right)\left(2020-x\right)}\)
Luôn có: \(0\le2\sqrt{\left(x-2019\right)\left(2020-x\right)}\le\left(x-2019\right)+\left(2020-x\right)\)
<=> \(1\le1+2\sqrt{\left(x-2019\right)\left(2020-x\right)}\le1+1\)
<=> \(1\le Q^2\le2\)
<=> \(1\le Q\le\sqrt{2}\)( do Q>0)
minQ=1 <=> \(\sqrt{\left(x-2019\right)\left(2020-x\right)}=0\)
<=> \(\left(x-2019\right)\left(2020-x\right)=0\)
<=> x=2019(tm) hoặc x=2020(t/m)
maxQ=\(\sqrt{2}\) <=> \(x-2019=2020-x\) <=> \(x=\frac{4039}{2}\) (tm)
\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)
\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)
\(\Rightarrow x-y=1\Rightarrow P=1\)
\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)
\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)
\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)
TXĐ: \(D=\left(-1;1\right)\)
\(B=\frac{2018x+2019\sqrt{1-x^2}+2020}{\sqrt{1-x^2}}\)
\(=\frac{2018x+2020}{\sqrt{1-x^2}}+2019\)
Đặt \(A=\frac{2018x+2020}{\sqrt{1-x^2}}>0\)vì \(-1< x< 1\)
=> \(\sqrt{1-x^2}.A=2018x+2020\)
=> \(\left(1-x^2\right)A^2=2018^2x^2+2.2018.2020x+2020^2\)
<=> \(\left(2018^2+A^2\right)x^2+2.2018.2020x+2020^2-A^2=0\)
pt trên có nghiệm <=> \(\Delta\ge0\)<=> \(\left(2018.2020\right)^2-\left(2018^2+A^2\right).\left(2020^2-A^2\right)\ge0\)
<=> \(A^4-\left(2020^2-2018^2\right)A^2\ge0\)
<=> \(A^2-8076\ge0\)
<=> \(A\ge\sqrt{8076}\)
"=" xảy ra <=> \(x=-\frac{1009}{1010}\left(tm\right)\)
Vậy GTNN của B = \(\sqrt{8076}+2019\) đạt tại \(x=-\frac{1009}{1010}\)
Ta có: \(\sqrt{x^2-2x+10}=\sqrt{x^2-2x+1+9}=\sqrt{\left(x-1\right)^2+9}\ge\sqrt{9}\ge3\)
\(\sqrt{x^2+4x+5}=\sqrt{x^2+4x+4+1}=\sqrt{\left(x+2\right)^2+1}\ge\sqrt{1}\ge1\)
\(\Rightarrow\) \(\sqrt{x^2-2x+10}+\sqrt{x^2+4x+5}\ge1+3\ge4\)
Vậy GTNN của biểu thức là 4
a: M=A:B
\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)
b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)
=>\(M=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)
Dấu = xảy ra khi (căn x+3)^2=16
=>căn x+3=4
=>x=1
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
Sửa đề: \(M=2019\sqrt{x-2}+2020\sqrt{10-y}\)
+Có: \(\sqrt{x-2}\ge với\forall x\\ \sqrt{10-y}\ge0với\forall x\\ \Rightarrow2019\sqrt{x-2}+2020\sqrt{10-y}\ge0\\ \Leftrightarrow M\ge0\)
+Dấu ''='' xảy ra khi
\(\sqrt{x-2}=0\\ \Leftrightarrow x=2\)
\(\sqrt{10-y}=0\\ \Leftrightarrow y=10\)
+Vậy \(M_{min}=0\) khi \(x=2,y=10\)
Đề không sai nha bạn