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

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

\(\Rightarrow1+x\sqrt{x^2+1}=\sqrt{x^2+1}+x\)

\(\Rightarrow1+x\sqrt{x^2+1}-\sqrt{x^2+1}-x=0\)

\(\Rightarrow-\left(x-1\right)+\left(x-1\right)\sqrt{x^2+1}=0\)

\(\Rightarrow\left(x-1\right)\left(\sqrt{x^2+1}-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2+1}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

\(a,2y^2-x+2xy=y+4\\ \Leftrightarrow2y\left(x+y\right)-\left(x+y\right)=4\\ \Leftrightarrow\left(2y-1\right)\left(x+y\right)=4=4\cdot1=\left(-4\right)\left(-1\right)=\left(-2\right)\left(-2\right)=2\cdot2\)

Vì \(x,y\in Z\Leftrightarrow2y-1\) lẻ 

\(\left\{{}\begin{matrix}2y-1=1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y-1=-1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)

Vậy PT có nghiệm \(\left(x;y\right)=\left\{\left(3;1\right);\left(4;0\right)\right\}\)

ĐKXĐ: \(x\ge-1\)

\(\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-6\sqrt{x+1}+9}=2\sqrt{x+1-2\sqrt{x+1}+1}\)

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

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

Ta có:

\(\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|\ge\left|\sqrt{x+1}+1+\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\)

Dấu "=" xảy ra khi và chỉ khi:

\(\sqrt{x+1}-3\ge0\Rightarrow x\ge8\)

Vậy nghiệm của pt là \(x\ge8\)

\(\sqrt{x+2\sqrt{x}+1}-\sqrt{x-2\sqrt{x}+1}=2\left(x\ge0\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow\sqrt{x}+1-\left|\sqrt{x}-1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1-\left(\sqrt{x}-1\right)=2,\forall\sqrt{x}-1\ge0\\\sqrt{x}+1-\left(1-\sqrt{x}\right)=2,\forall\sqrt{x}-1< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0\sqrt{x}=0,\forall x\ge1\\\sqrt{x}=1,\forall x< 1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in R,x\ge1\\x=1,x< 1\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x\in R,x\ge1\)

\(\sqrt{x-2+2\sqrt{x+1}}+\sqrt{x+10+6\sqrt{x+1}}=2\sqrt{x+2+2\sqrt{x+1}}\)

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

Đặt \(y=\sqrt{x+1}\left(y\ge0\right)\)PT đã cho trở thành

\(y+1+\left|y-3\right|=2\left|y-1\right|\)

Nếu \(0\le y\le1:y+1+3-y=2-2y\Leftrightarrow y=-1\)(loại)

Nếu \(1\le y\le3:y+1+3-y=2y-2\Leftrightarrow y=3\)

Nếu y>3: y+1-y-3=2y-2 (vô nghiệm)

Với y=3 <=> x+1=9 <=> x=8

Vậy pt có 1 nghiệm x=8

bài này bạn dùng cách nhân với 1 lượng liên hợp:

<=> \(\frac{\sqrt{X+3}-\sqrt{X+2}}{x+3-x-2}\)+\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}\)+\(\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\)=1

<=>\(\sqrt{x+3}-\sqrt{x}=1\)

<=> \(\sqrt{x+3}=1+\sqrt{x}\)

Tới đây bình phương hai vế, ta có:

x+3 =1+2\(\sqrt{x}\)+x

<=> 2\(\sqrt{x}\)=2 <=> X=1

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{1}{2}\left(x+3\right)\)

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

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

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{1}{2}\left(x+3\right)\)

TH1: \(x>=2\)

PT sẽ tương đương với \(\sqrt{x-1}+1+\sqrt{x-1}-1=\dfrac{1}{2}\left(x+3\right)\)

=>\(2\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)

=>\(4\sqrt{x-1}=x+3\)

=>\(\sqrt{16x-16}=x+3\)

=>x>=-3 và (x+3)^2=16x-16

=>x>=-3 và x^2+6x+9-16x+16=0

=>x>=-3 và x^2-7x+25=0

=>Loại

TH2: 1<=x<2

PT sẽ là \(\sqrt{x-1}+1+1-\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)

=>1/2(x+3)=2

=>x+3=4

=>x=1(nhận)

ĐKXĐ: \(x\ge1\)

\(x-1+\sqrt{5+\sqrt{x-1}}=5\)

Đặt \(\sqrt{x-1}=t\ge0\)

\(\Rightarrow t^2+\sqrt{t+5}=5\)

Đặt \(\sqrt{t+5}=u>0\Rightarrow u^2-t=5\)

\(\Rightarrow t^2+u=u^2-t\Leftrightarrow t^2-u^2+t+u=0\)

\(\Leftrightarrow\left(t+u\right)\left(t-u+1\right)=0\)

\(\Leftrightarrow t-u+1=0\) (do \(t>0;u>0\Rightarrow t+u>0\))

\(\Leftrightarrow t+1=\sqrt{t+5}\)

\(\Leftrightarrow t^2+2t+1=t+5\Leftrightarrow t^2+t-4=0\)

\(\Rightarrow t=\dfrac{-1+\sqrt{17}}{2}\)

\(\Rightarrow x=t^2+1=\dfrac{11-\sqrt{17}}{2}\)