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

no biet

i don't known

a) \(\sqrt{4-5x}=12\)

ĐK : x ≤ 4/5

Bình phương hai vế

⇔ \(4-5x=144\)

⇔ \(-5x=140\)

⇔ \(x=-28\)( tm )

b) \(\sqrt{1-4x+4x^2}=5\)

⇔ \(\sqrt{\left(1-2x\right)^2}=5\)

⇔ \(\left|1-2x\right|=5\)

⇔ \(\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐK : x ≥ -5

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(\left|2\right|\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot\left|3\right|\sqrt{x+5}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\)( tm )

d)\(\sqrt{x-2}\le3\)

ĐK : x ≥ 2

⇔ \(x-2\le9\)

⇔ \(x\le11\)

Kết hợp với điều kiện => Nghiệm của bpt là 2 ≤ x ≤ 11

\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy pt vô nghiệm

\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)

\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(\sqrt{x-5}=\sqrt{3-x}\)

⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)

⇔\(x-5=3-x\)

⇔\(x=4\)

b) \(\sqrt{4-5x}=\sqrt{2-5x}\)

⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)

⇔\(4-5x=2-5x\)

⇔\(2=0\) (Vô lí)

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

\(1-\sqrt{2}x\) nha

\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)

\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)

\(\Leftrightarrow4x^2+4x-1=0\)

\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)

\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)

\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)

\(M=1+8+1=10\)