đkxđ : x+  2018 >= 0

<=> x >= -2018

vậy_

\(ĐKXĐ:\sqrt{x+2018}\ne0\)

\(\Rightarrow x+2018\ne0\)

\(\Rightarrow x\ne-2018\)

#H

Bài làm:
\(N=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(N=\frac{-4}{1+\sqrt{5}}\cdot\frac{4}{1-\sqrt{5}}\)

\(N=\frac{-16}{1-5}=\frac{-16}{-4}=4\)

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

Để M nguyên thì \(\sqrt{x}-2\)phải là ước của 2 hay 

\(\sqrt{x}-2=-2;-1;1;2\)

\(\Rightarrow x=0;1;9;16\)

\(\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\frac{2}{\sqrt{x}-2}\)ĐK : \(x\ne4\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Tương tự bài a mới làm 

ĐKXĐ : \(\sqrt{x}+1\ge1\)(1)

\(C\inℤ\Leftrightarrow\sqrt{x}+1\inƯ\left(-4\right)\)(2) 

Từ (1) và (2) \(\Rightarrow\sqrt{x}+1\in\left\{1;2;4;\right\}\)

\(\Rightarrow x\in\left\{0;1;9\right\}\)

sai đề bn ơi

không sai đâu nha bn, mình làm đc rồi

\(\sqrt{x^2-6x+9}=\sqrt{4x^2+4x+2}\)

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

\(\left(x-3\right)^2=4x^2+4x+2\)

\(\left(x-3\right)^2-\left(2x+1\right)^2-1=0\)

\(\left(x-3-2x-1\right)\left(x-3+2x+1\right)-1=0\)

\(\left(-x-4\right)\left(3x-2\right)-1=0\)

\(-3x^2-12x+2x+8-1=0\)

\(-3x^2-10x+7=0\)

\(\sqrt{\Delta}=\sqrt{\left(-10\right)^2-4\left(-3\right)7}=2\sqrt{46}\)

\(\orbr{\begin{cases}x_1=\frac{10+2\sqrt{46}}{-6}=\frac{-5-\sqrt{46}}{3}\left(TM\right)\\x_2=\frac{10-2\sqrt{46}}{-6}=\frac{\sqrt{46}-5}{3}\left(TM\right)\end{cases}}\)

\(\sqrt{x^2-6x+9}=\sqrt{4x^2+4x+2}\)

\(\Leftrightarrow x^2-6x+9=4x^2+4x+2\Leftrightarrow3x^2+10x-7=0\)

\(\Delta'=25-\left(-7\right).3=46>0\)

pt có 2 nghiệm pb 

\(x_1=\frac{-5-\sqrt{46}}{3};x_2=\frac{-5+\sqrt{46}}{3}\)

\(\sqrt[23]{\frac{345}{80.78}}\)

bạn đánh dấu cho mình nghe

a, \(\sqrt{9\left(x+2\right)^2}=6\Leftrightarrow3\sqrt{\left(x+2\right)^2}=6\Leftrightarrow\left|x+2\right|=2\)

TH1 : \(x+2=2\Leftrightarrow x=0\)

TH2 : \(x+2=-2\Leftrightarrow x=-4\)

b, \(3x+\sqrt{4x^2-4x+1}=4\Leftrightarrow\sqrt{\left(2x-1\right)^2}=4-3x\Leftrightarrow\left|2x-1\right|=4-3x\)

ĐK : \(4-3x\ge0\Leftrightarrow-3x\ge-4\Leftrightarrow x\le\frac{4}{3}\)

TH1 : \(2x-1=4-3x\Leftrightarrow5x=5\Leftrightarrow x=1\)(tm)

TH2 : \(2x-1=3x-4\Leftrightarrow x=3\)(ktm)