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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
\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)
\(\Leftrightarrow4x^2+4x-1=0\)
\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)
\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)
\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
\(M=1+8+1=10\)
b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)
\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)
\(\frac{1}{pt}\)=\(\sqrt{x}+\sqrt{2x+3}=\frac{1}{\sqrt{3}}\left(\sqrt{4x-3}+\sqrt{5x-6}\right)\)
=>\(\frac{x-2x-3}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{4x-3-5x-6}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)
=>\(\frac{3-x}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{3-x}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)
=>\(\sqrt{x}-\sqrt{2x-3}=\sqrt{3}\left(\sqrt{4x-3}-\sqrt{5x+6}\right)\)
=>\(\frac{3-x}{\sqrt{x}+\sqrt{2x-3}}=\sqrt{3}\left(\frac{3-x}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)
=>\(\left(3-x\right)\left(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\right)\)=0
=>3-x=0=>x=3
hoặc\(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)=0
Em mới học lớp 7