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1: ĐKXĐ: x>1/2
=>\(\dfrac{x}{\sqrt{2x-1}}+\dfrac{x}{\sqrt[4]{4x-3}}=2\)
x^2-2x+1>=0
=>x^2>=2x-1
=>\(\dfrac{x}{\sqrt{2x-1}}>=1\)
Dấu = xảy ra khi x=1
(x^2-2x+1)(x^2+2x+3)>=0
=>x^4-4x+3>=0
=>x^4>=4x-3
=>\(\dfrac{x}{\sqrt[4]{4x-3}}>=1\)
=>VT>=2
Dấu = xảy ra khi x=1
2: 4x-1=x+x+2x-1
5x-2=x+2x-1+2x-1
\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}\right)\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)>=9\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)
\(\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)^2< =3\left(4x-1\right)\)
=>\(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}< =\sqrt{3\left(4x-1\right)}\)
=>\(\dfrac{2}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{4x-1}}\)
Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\sqrt{3}\left(\dfrac{1}{\sqrt{4x-1}}+\dfrac{1}{\sqrt{5x-2}}\right)\)
Dấu = xảy ra khi x=1
Bạn ghi lộn đề rồi \(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}\) chứ không phải \(\left(\dfrac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)^{2014}\)
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2-1}\right)}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\left|\sqrt{2}-1\right|}{2}=\dfrac{\sqrt{2}-1}{2}\)
Vậy ta có \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)Ta lại có \(\left(4x^5+4x^4-x^3+1\right)^{19}=\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=\left(x^3.0+1\right)^{19}=1^{19}=1\)(1)
\(\left(\sqrt{4x^5+4x^4-5x^3+5x+3}\right)^3=\left(\sqrt{4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1+4}\right)^3=\left(\sqrt{x^3\left(4x^2+4x-1\right)-x^2\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\right)^3=\left(\sqrt{x^3.0+x^2.0+0+4}\right)^3=\left(\sqrt{4}\right)^3=2^3=8\left(2\right)\)
\(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}=\left[\dfrac{1-\sqrt{2}.\dfrac{\sqrt{2}-1}{\sqrt{2}}}{\sqrt{2.\dfrac{3-2\sqrt{2}}{4}+\sqrt{2}-1}}\right]^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}}{2}+\sqrt{2}-1}}\right)^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}+2\sqrt{2}-2}{2}}}\right)^{2014}=\left(\dfrac{\dfrac{\dfrac{1}{\sqrt{2}}}{1}}{\sqrt{2}}\right)^{2014}=1^{2014}=1\left(3\right)\)
Cộng (1),(2),(3) theo vế ta được A=1+8+1=10
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=10
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{1}{2}.\left|\sqrt{2}-1\right|=\dfrac{\sqrt{2}-1}{2}\)
Vậy \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)
Ta lại có \(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2017}+2019=\left(4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1-1\right)^{2017}+2019=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)-1\right]^{2017}+2019=\left(x^3.0-x.0+0-1\right)^{2017}+2019=\left(-1\right)^{2017}+2019=-1+2019=2018\)
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=2018
Ta có:
x = \(\frac{1}{2}\)\(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
= \(\frac{1}{2}\)\(\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{1}}\)
= \(\frac{1}{2}\)(\(\sqrt{2}\)-1)
=> 2x = \(\sqrt{2}\)-1
=> (2x)2= ( \(\sqrt{2}\)-1)2
=> 4x2= 2-2\(\sqrt{2}\)+1
=> 4x2= -2( \(\sqrt{2}\)-1)+1
=> 4x2= -4x +1 => 4x2+4x-1=0
Lại có:
A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19
= [ x3( 4x2+4x-1) +1]19
=1
A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3
= (\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\))3
= 23=8
A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)
= \(\sqrt{2}\)- \(\sqrt{2}\)\(\sqrt{1-\sqrt{2}}\)
Cộng 3 số vào ta được A
b)\(x+\dfrac{4}{x+2}=3\) (ĐKXĐ: \(x\ne-2\))
\(\Leftrightarrow x+\dfrac{4}{x+2}-3=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)+4-3\left(x+2\right)}{x+2}=0\\ \Leftrightarrow\dfrac{x^2+2x+4-3x-6}{x+2}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{x+2}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x=\dfrac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4.1.\left(-2\right)}}{2.1}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{1+8}}{2}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{9}}{2}\\ \Leftrightarrow x=\dfrac{1\pm3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+3}{2}\\x=\dfrac{1-3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy...