giê ơt nha bn

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

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

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

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

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

=2.

dấu = khi và chỉ khi \(\left(\sqrt{x-1}+1\right).\left(1-\sqrt{x-1}\right)=0\)

=0 nha bn

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4+\dfrac{1}{x^2}}-\sqrt{\dfrac{1}{x}+\dfrac{5}{x^2}}}{2-\dfrac{7}{x}}=1\)

ĐK: \(x\ge1\)

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

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

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

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

Đẳng thức xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\)

\(\Leftrightarrow1\le x\le2\)

ĐK: x\(\ge\)2

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

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

\(E=\dfrac{\left|\sqrt{x+2}+\sqrt{x-2}\right|}{\sqrt{x^2-4}+x+2}\)

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

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

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

Thế x=2(\(\sqrt{3}+1\))=\(2\sqrt{3}+2\) vào E:

=>\(E=\dfrac{1}{\sqrt{2\sqrt{3}+4}}\)

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ge0\\x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có : \(B=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{2\sqrt{x}}\)

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

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

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

Vậy ....

Đặt \(\sqrt{x+1}=t\left(t\ge0\right)\Rightarrow x^2=t^2-1\)

\(pt\Leftrightarrow t^2-1+t=1\)

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

\(\Leftrightarrow\orbr{\begin{cases}t=-2\left(loại\right)\\t=1\end{cases}}\)

Với \(t=1\Leftrightarrow\sqrt{x+1}=1\Leftrightarrow x+1=1\Leftrightarrow x=0\)

KL: \(x=0\)

không dùng ẩn phụ được không ạ ?

\(x^2+\sqrt{x+1}=1\left(đk:x\ge-1\right)\)\(< =>x^2+\sqrt{x+1}-1=0\)

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

\(< =>x=0\)và xử lí phần trong ngoặc là ok

\(x^2+\sqrt{x+1}=1\)  ( 1 ) 

Đặt \(t=\sqrt{x+1}\left(ĐK:t\ge0;x\ge-1\right)\)

\(t^2=x+1\) 

\(t^2-1=x\)         

\(\left(t^2-1\right)^2+t=1\) 

\(t^4-2t^2+1+t-1=0\)  

\(t^4-2t^2+t=0\) 

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

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

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

t = 0 ( nhận ) hoặc t = 1 (nhận ) hoặc \(t=\frac{-1+\sqrt{5}}{2}\)   ( nhận ) hoặc \(x=\frac{-1-\sqrt{5}}{2}\) ( loại ) 

Với t = 0 

\(\sqrt{x+1}=0\)

 \(x+1=0;x=-1\) 

t = 1 

\(\sqrt{x+1}=1\) 

\(x+1=1;x=0\) 

\(t=\frac{-1+\sqrt{5}}{2}\) 

\(\sqrt{x+1}=\frac{-1+\sqrt{5}}{2}\) 

\(x+1=\frac{3-\sqrt{5}}{2};x=\frac{1-\sqrt{5}}{2}\) 

vậy \(x=-1\) hoặc \(x=0\) hoặc \(x=\frac{1-\sqrt{5}}{2}\) là nghiệm của phương trình 

\(\left\{{}\begin{matrix}x+2y=5m-1\\-2x+y=2\end{matrix}\right.< =>\left\{{}\begin{matrix}2x+4y=10m-2\\-2x+y=2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}5y=10m\\-2x+y=2\end{matrix}\right.< =>\left\{{}\begin{matrix}y=2m\\x=m-1\end{matrix}\right.\)

=>\(\sqrt{x}+\sqrt{y}=\sqrt{2}\left(1\right)\) 

=>\(\sqrt{m-1}+\sqrt{2m}=\sqrt{2}\) (\(m\ge1\))

\(< =>\left(\sqrt{m-1}\right)^2=|\left(\sqrt{2}-\sqrt{2m}\right)^2|\)

<=>\(m-1=\left[\sqrt{2}.\left(1-\sqrt{m}\right)\right]^2< =>m-1=|2.\left(1-\sqrt{m}\right)^2|\)

<=>\(m-1=|2\left(1-2\sqrt{m}+m\right)|=\left|2-4\sqrt{m}+2m\right|\)

với \(\left|2-4\sqrt{m}+2m\right|=2-4\sqrt{m}+2m< =>m\le1\)

ta có pt:

<=>\(m-1-2+4\sqrt{m}-2m=0\)

\(< =>-m+4\sqrt{m}-3=0< =>-\left(m-4\sqrt{m}+3\right)=0\)

<=>\(m-4\sqrt{m}+3=0< =>\left(\sqrt{m}-3\right)\left(\sqrt{m}-1\right)=0\)

<=>\(\left[{}\begin{matrix}\sqrt{m}-3=0\\\sqrt{m}-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}m=9\left(loai\right)\\m=1\left(TM\right)\end{matrix}\right.\) 

nếu \(|2-4\sqrt{m}+2m|=-2+4\sqrt{m}-2m< =>m\ge1\)

=>\(-2+4\sqrt{m}-2m=m-1< =>3m-4\sqrt{m}+1=0\)

<=>\(3\left(m-2.\dfrac{2}{3}\sqrt{m}+\dfrac{1}{3}\right)=3\left(m-2.\dfrac{2}{3}\sqrt{m}+\dfrac{4}{9}-\dfrac{4}{9}+\dfrac{1}{3}\right)=0\)

<=>\(\left(\sqrt{m}-1\right)\left(\sqrt{m}-\dfrac{1}{3}\right)=0\)=>\(\left[{}\begin{matrix}\sqrt{m}-1=0\\\sqrt{m}-\dfrac{1}{3}=0\end{matrix}\right.< =>\left\{{}\begin{matrix}m=1\left(TM\right)\\m=\dfrac{1}{3}\left(loai\right)\end{matrix}\right.\)

vậy m=1 thì pt đã cho có 2 nghiệm (x,y) thỏa mãn

\(\sqrt{x}+\sqrt{y}=\sqrt{2}\)

chỗ cuối sửa thành x=1/9 (loại ) hộ :((