Điều kiện : \(x\ne1\)

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

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

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

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

+) Với \(-3< x< 1\) thì VP < 0 => Phương trình vô nghiệm

+) Với \(\left[{}\begin{matrix}x\le-3\\x>1\end{matrix}\right.\) thì VP > 0 , lúc này ta có phương trình : 

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

\(\Leftrightarrow x^2+2x+3=\left(x^2+2x+1-4\right)^2\)

\(\Leftrightarrow x^2+2x+3=\left(x+1\right)^4-8\left(x+1\right)^2+16\)

\(\Leftrightarrow\left(x+1\right)^2+2=\left(x+1\right)^4-8\left(x+1\right)^2+16\)

\(\Leftrightarrow\left(x+1\right)^4-9\left(x+1\right)^2+14=0\)

Đặt \(\left(x+1\right)^2=t\left(t>0\right)\) , ta có : 

\(t^2-9t+14=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=7\\t=2\end{matrix}\right.\)

+) \(t=7\Rightarrow\left(x+1\right)^2=7\Leftrightarrow x=\sqrt{7}-1\left(tm\right)\)

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

Vậy \(x=\sqrt{7}-1\)

Điều kiện : x > 0 

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

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

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

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

+) \(1-x< 0\Leftrightarrow x>1\) thì phương trình vô nghiệm do \(VT>0\)

+) \(1-x>0\Leftrightarrow x< 1\), kết hợp với x > 0 

\(\Rightarrow0< x< 1\)

Lúc này ta được phương trình :

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

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

\(\Leftrightarrow x^2-x^2+x+2x-1=0\)

\(\Leftrightarrow3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)

Vậy...............

\(Đk:\left\{{}\begin{matrix}x>0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ge-1\end{matrix}\right.\Leftrightarrow x>0\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x\left(x+1\right)=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2+x=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

- Vậy \(S=\left\{\dfrac{1}{3}\right\}\)

`17/13 - 106/111 + (-5/111)`

`=17/13 - 106/111 - 5/111`

`=17/13+(-106/111-5/111)`

`=17/13 + (-111/111)`

`=17/13 - 1`

`=17/13-13/13`

`=4/13`

=> [17/13-106/111] + (-5/111)
=> 17/13-509/1443
=>4/13

\(Đk:2x+3\ge0\Leftrightarrow x\ge-\dfrac{3}{2}\)

\(x^2+4x+5=2\sqrt{2x+3}\)

\(\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

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

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

\(\Leftrightarrow x=-1\left(nhận\right)\)

- Vậy \(S=\left\{-1\right\}\)

Điều kiện : \(x\ge-\dfrac{3}{2}\)

\(x^2+4x+5=2\sqrt{2x+3}\)

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

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

Đặt \(x+2=t\left(t\ge\dfrac{1}{2}\right)\)

Ta có phương trình :

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

\(\Leftrightarrow\left(t^2+1\right)^2=4\left(2t-1\right)\)

\(\Leftrightarrow t^4+2t^2+1-8t+4=0\)

\(\Leftrightarrow t^4+2t^2-8t+5=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow x+2=1\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy \(\Leftrightarrow x=-1\left(tm\right)\)

 300:6=50
số quyển vở mua thêm là 
 
=> 6+16=22 quyển

22

\(\sqrt{5-x}=2x-7\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-7\ge0\\5-x=\left(2x-7\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\5-x=4x^2-28x+49\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\4x^2-27x+44=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\4x^2-16x-11x+44=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\4x\left(x-4\right)-11\left(x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\\left(x-4\right)\left(4x-11\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{7}{2}\\\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{11}{4}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

- Thử lại, ta có \(x=4\) là nghiệm của phương trình trên.

- Vậy \(S=\left\{4\right\}\)

Điều kiện : \(x\le5\)

+) Với \(2x-7< 0\Leftrightarrow x< \dfrac{7}{2}\Rightarrow VP< 0\)

=> Phương trình vô nghiệm 

+) Với \(2x-7>0\Leftrightarrow x>\dfrac{7}{2}\Rightarrow VP>0\)

Ta có phương trình :

\(5-x=\left(2x-7\right)^2\)

\(\Leftrightarrow5-x=4x^2-28x+49\)

\(\Leftrightarrow4x^2-27x+44=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{11}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy x = 4

\(H=\dfrac{6-8x}{x^2+1}\)

<=> Hx² + H = 6 - 8x 

<=> Hx² + 8x + H - 6 = 0 (1) 

Phương trình (1) có nghiệm khi 

\(\Delta=8^2-4H\left(H-6\right)\ge0\)

<=> \(H^2-6H-16\le0\)

<=> \(\left(H-8\right)\left(H+2\right)\le0\)

\(\Leftrightarrow-2\le H\le8\) 

=> Min H = -2 

Dấu "=" xảy ra khi x = 2