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Bài 1 : Ta có :
\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)
\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)
\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)
\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)
\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)
\(=2\sqrt{6x-1}\)
\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)
Thay \(x=4+\sqrt{10}\) vào A ta được :
\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)
\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)
\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)
Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)
Giả sử đề bạn là 2012 thì mình làm nhé.
\(x^4+\sqrt{x^2+2012}=2012\)
\(\Leftrightarrow\left(x^4+x^2+\dfrac{1}{4}\right)=\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{2}\right)^2=\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow x^2+\dfrac{1}{2}=\sqrt{x^2+2012}-\dfrac{1}{2}\)
\(\Leftrightarrow\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)=2011,25\)
\(\Leftrightarrow\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2=2011,25\)
Tới đây thì đơn giản rồi. b làm tiếp nhé
1/ Đặt \(\hept{\begin{cases}\sqrt{x-2013}=a\\\sqrt{x-2014}=b\end{cases}}\)
Thì ta có:
\(\frac{\sqrt{x-2013}}{x+2}+\frac{\sqrt{x-2014}}{x}=\frac{a}{a^2+2015}+\frac{b}{b^2+2014}\)
\(\le\frac{a}{2a\sqrt{2015}}+\frac{b}{2b\sqrt{2014}}=\frac{1}{2\sqrt{2015}}+\frac{1}{2\sqrt{2014}}\)
2/ \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)\)
\(=\frac{3}{4}\)
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow1\ge4xy\Leftrightarrow xy\le\frac{1}{4}\)(1)
\(\left(x-y\right)^2\ge0\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge2\Leftrightarrow x+y\ge\sqrt{2}\)
Từ phần a ta có \(x+y\le\sqrt{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2\)
\(\le\left(1+1\right)\left(2\left(x+y\right)+2\right)\)
\(=2\cdot\left(2\left(x+y\right)+2\right)\le2\cdot\left(2\sqrt{2}+2\right)\)
\(=4\sqrt{2}+4=VP^2\)
Suy ra \(VT\ge VP\) (ĐPCM)
\(x=\dfrac{\sqrt{\sqrt{5}-2}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+1\right)}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(x=\dfrac{1+\sqrt{5}-2}{\sqrt{3-\sqrt{5}}}-\left(\sqrt{2}-1\right)=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{6-2\sqrt{5}}}-\left(\sqrt{2}-1\right)\)
\(x=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{\left(\sqrt{5}-1\right)^2}}-\sqrt{2}+1=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{2}+1=1\)
\(\Rightarrow x^{2012}+2x^{2013}+3x^{2014}=1^{2012}+2.1^{2013}+3.1^{2014}=6\)
Sửa đề: \(\sqrt{2010}-2\sqrt{2012}+\sqrt{2014}< 0\)
Ta có: \(\left(\sqrt{2010}+\sqrt{2014}\right)^2\)
\(=2010+2\sqrt{2010\cdot2014}+2014\)
\(=4024+2\sqrt{\left(2012-2\right)\left(2012+2\right)}\)
\(=2\cdot2012+2\sqrt{2012^2-2^2}\)
\(< 2\cdot2012+2\cdot\sqrt{2012^2}=2\cdot2012+2\cdot2012\)
\(=4\cdot2012=\left(2\sqrt{2012}\right)^2\)
\(\Rightarrow\sqrt{2010}+\sqrt{2014}< 2\sqrt{2012}\)
\(\Leftrightarrow\sqrt{2010}-2\sqrt{2012}+\sqrt{2014}< 0\)
Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(2014-x+x-2012\right)\left(1^2+1^2\right)\ge\left(\sqrt{2014-x}+\sqrt{x-2012}\right)^2\)
\(\Leftrightarrow\left(\sqrt{2014-x}+\sqrt{x-2012}\right)^2\le4\left(2012\le x\le2014\right)\)
\(\Leftrightarrow\sqrt{2014-x}+\sqrt{x-2012}\le2\)
\("="\Leftrightarrow x=2013\left(TM\right)\)