ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

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Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

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

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

<=> ( x - 2 )( 2x - 5 ) = 0

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

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

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

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

1: 5(2-3x)(x-2)=3(1-3x)

=>5(3x-2)(x-2)=3(3x-1)

=>\(5\cdot\left(3x^2-6x-2x+4\right)=9x-3\)

=>\(15x^2-40x+20-9x+3=0\)

=>\(15x^2-49x+23=0\)

\(\Delta=49^2-4\cdot15\cdot23=1021>0\)

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

\(\left[\begin{array}{l}x=\frac{49-\sqrt{1021}}{2\cdot15}=\frac{49-\sqrt{1021}}{30}\\ x=\frac{49+\sqrt{1021}}{2\cdot15}=\frac{49+\sqrt{1021}}{30}\end{array}\right.\)

2: \(4x^2+4x+1=0\)

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

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\)

3: \(4x^2-9=0\)

=>\(4x^2=9\)

=>\(x^2=\frac94\)

=>\(\left[\begin{array}{l}x=\frac32\\ x=-\frac32\end{array}\right.\)

4: \(5x^2-10x=0\)

=>5x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

5: \(x^2-3x=-2\)

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

=>(x-1)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

6: |x-5|-3=0

=>|x-5|=3

=>\(\left[\begin{array}{l}x-5=3\\ x-5=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=2\end{array}\right.\)

lập Vi-ét mà tính

muốn kiểm tra đáp án nên hỏi, chứ ai chả biết ?

\(\Leftrightarrow\left(x^2+2\right)\sqrt{x^2+x+1}-2\left(x^2+2\right)+x^3-x^2-5x+6=0\)

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

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

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

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

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

Đặt \(\sqrt{x^2+x+1}=t>0\Rightarrow x^2=t^2-x-1\)

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

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

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

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

\(\Leftrightarrow t=3-x\)

\(\Leftrightarrow\sqrt{x^2+x+1}=3-x\) (\(x\le3\))

\(\Leftrightarrow x^2+x+1=x^2-6x+9\)

\(\Leftrightarrow x=\dfrac{8}{7}\)