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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow2x+1=\sqrt{2}\)

\(\Leftrightarrow4x^2+4x-1=0\)
Giờ thế vô A đi

ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

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

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Bài 1:

ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$

$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$

Coi đây là PT bậc 2 ẩn $x$

$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:

$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:

$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$