a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)

=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)

\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)

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

\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)

b: \(5x^2-3x+1=2x+31\)

=>\(5x^2-3x+1-2x-31=0\)

=>\(5x^2-5x-30=0\)

=>\(x^2-x-6=0\)

=>(x-3)(x+2)=0

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

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

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

=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)

\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)

\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0

Do đó, phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)

Nhìn không đủ chán rồi không dám động vào

Viết đề kiểu gì v @@

À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)

a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)   (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )

Khi đó pt :

<=> 7+a =3 + \(\sqrt{5}\)

<=> 4+a = \(\sqrt{5}\)

<=> (4+a)\(^2\) = 5

<=> 16 + 8a + a\(^2\) = 5

<=>a\(^2\) + 8a+ 11 = 0

<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại) 

Vậy Pt vô nghiệm.

b) \(\sqrt{3x^2-4x}\) = 2x-3

<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9 

<=> x\(^2\)-8x+9 = 0

<=> x=1 , x=9 

Vậy S={1;9} 

c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2

<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)

<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)

<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)

<=> x=7,x=5

Vậy x=5 hoặc x=7

a, \(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy ...

b, ĐKXĐ : \(x\ge-1\)

\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=-\dfrac{5}{3}\)

Vậy phương trình vô nghiệm

a)Pt \(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge-1\)

Pt\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow-3\sqrt{x+1}=5\) (vô nghiệm)

Vậy...

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

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy S={\(\dfrac{5}{2};-\dfrac{3}{2}\)}