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

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

P = \(\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{1+2x-4x-2\sqrt{x}-1+4x}\)

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

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

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

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

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

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

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

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

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

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

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

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: \(x\ge1\)

Nhân cả tử và mẫu của vế phải với liên hợp của nó ta được:

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

Đặt \(\sqrt{x+1}+\sqrt{x-1}=a>0\)

\(\Rightarrow a^2-3=\frac{a}{2}\Rightarrow2a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+1}+\sqrt{x-1}=2\)

\(\Leftrightarrow x+\sqrt{x^2-1}=2\)

\(\Leftrightarrow\sqrt{x^2-1}=2-x\) (\(x\le2\))

\(\Leftrightarrow x^2-1=x^2-4x+4\)

\(\Rightarrow x=\frac{5}{4}\)

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)