a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+1-3\right)+ 4\left(x^2+x+1-3\right)=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+5\right)=0\)

\(\Leftrightarrow x^2+x+4=0\) hay \(x^2+x-2=0\)

\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{15}{4}=0\) hay \(x^2-x+2x-2=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (pt vô nghiệm) hay\(x\left(x-1\right)+2\left(x-1\right)=0\)

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

\(\Leftrightarrow x=1\) hay \(x=-2\)

-Vậy \(S=\left\{1;-2\right\}\)

b) \(x^3+5x^2-10x-8=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+4x-8=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+7x+4\right)=0\)

\(\Leftrightarrow x=2\) hay \(x^2+2.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{7}{2}-\dfrac{\sqrt{33}}{2}\right)=0\)

\(\Leftrightarrow x=2\) hay \(x=\dfrac{-7-\sqrt{33}}{2}\) hay \(x=\dfrac{-7+\sqrt{33}}{2}\)

-Vậy \(S=\left\{2;\dfrac{-7-\sqrt{33}}{2};\dfrac{-7+\sqrt{33}}{2}\right\}\)

a) \(|2x+1|=|x-3|\)

\(\Leftrightarrow|2x+1|-|x-3|=0\)

Lập bảng xét dấu :

Nếu \(x< \frac{-1}{2}\) thì \(|2x+1|=-2x-1\)

                                    \(|x-3|=3-x\)

\(pt\Leftrightarrow\left(-2x-1\right)-\left(3-x\right)=0\)

\(\Leftrightarrow-2x-1-3+x=0\)

\(\Leftrightarrow-x=4\)

\(\Leftrightarrow x=-4\left(tm\right)\)

Nếu  \(\frac{-1}{2}\le x\le3\) thì \(|2x+1|=2x+1\)

                                               \(|x-3|=3-x\)

\(pt\Leftrightarrow\left(2x+1\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x+1-3+x=0\)

\(\Leftrightarrow3x-2=0\)

\(x=\frac{2}{3}\left(tm\right)\)

Nếu  \(x>3\) thì \(|2x+1|=2x+1\) 

                               \(|x-3|=x-3\)

\(pt\Leftrightarrow\left(2x+1\right)-\left(x-3\right)=0\)

\(\Leftrightarrow2x+1-x+3=0\)

\(\Leftrightarrow x=-4\) ( loại )

\(x^4+x^2+6x-8=0\)

\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(x^2-6x+9\right)=0\)

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

Mà \(\left(x^2+1\right)^2\ge0\forall x\)

      \(\left(x-3\right)^2\ge0\forall x\)

Dấu bằng xảy ra khi :

\(\hept{\begin{cases}x^2+1=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=-1\\x=3\end{cases}}\)

Lại có \(x^2\ge0\forall x\)

\(\Leftrightarrow x^2=-1\) ( vô lí )

Vậy phương trình có tập nghiệm \(S=\left\{3\right\}\)

Tham khảo bài này :

(3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy x = -1/3 hoặc x = -5

\(a,x^2+10x+25-4x\left(x+5\right)=0.\)

\(\Leftrightarrow\left(x+5\right)^2-4x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(5-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\5-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

\(b,\left(4x-5\right)^2-2\left(16x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-5\right)^2-2\left(4x+5\right)\left(4x-5\right)=0\)

\(\Leftrightarrow-\left(4x-5\right)\left(4x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-5=0\\4x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}}\)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>9x^2-23x=x^2-5x+6

=>8x^2-18x-6=0

=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)

b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>216x+18=275x-100

=>-59x=-118

=>x=2

2\(x^2\) - 5 \(\sqrt{x^2-5x+7}\) = 10\(x\) - 17 Đk \(x^2\) - 5\(x\) + 7  ≥ 0

\(x^2\) - 2.\(\dfrac{5}{2}\)\(x\) + \(\dfrac{25}{4}\) + \(\dfrac{3}{4}\) = (\(x\) - \(\dfrac{5}{2}\))² + \(\dfrac{3}{4}\) > 0 ∀ \(x\)

ta có: 2\(x^2\) - 5\(\sqrt{x^2-5x+7}\) = 10\(x\) - 17

2\(x^2\) - 5\(\sqrt{x^2-5x+7}\) - 10\(x\) + 17 = 0

(2\(x^2\) - 10\(x\) + 14)  -  5\(\sqrt{x^2-5x+7}\) + 3 = 0

2.(\(x^2\) - 5\(x\) + 7) - 5.\(\sqrt{x^2-5x+7}\) + 3 = 0

Đặt \(\sqrt{x^2-5x+7}\) = y > 0 ta có: 

2y² - 5y + 3  = 0

2 + (-5) + 3 = 0

⇒ y₁= 1; y₂ =  \(\dfrac{3}{2}\) 

TH1 y = 1 ⇒ \(\sqrt{x^2-5x+7}\)  = 1

⇒ \(x^2\) - 5\(x\) + 7  = 1

    \(x^2\) - 5\(x\) + 6 = 0

     \(\Delta\) = 25 -  24 = 49

    \(x_1\) = \(\dfrac{-\left(-5\right)+\sqrt{1}}{2}\) =  3;

    \(x_2\) =  \(\dfrac{-\left(-5\right)-\sqrt{1}}{2}\)  = 2;

TH2  y = \(\dfrac{3}{2}\)

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

         \(x^2\) - 5\(x\) + 7 = \(\dfrac{9}{4}\)

         4\(x^2\) - 20\(x\) + 28 = 9

          4\(x^2\) - 20\(x\) + 19 = 0

           \(\Delta'\) = 10² - 4.19

          \(\Delta'\) = 24

           \(x_1\) = \(\dfrac{-\left(-10\right)+\sqrt{24}}{4}\) = \(\dfrac{10+\sqrt{24}}{4}\)

           \(x_2\) = \(\dfrac{-\left(-10\right)-\sqrt{24}}{4}\) = \(\dfrac{10-\sqrt{24}}{4}\)

            8 - 5\(\sqrt{6}\)

Từ các lập luận trên kết luận phương trình có tập nghiệm là:

S = {8 - 5\(\sqrt{6}\); 2 ; 3; 8 + 5\(\sqrt{6}\)}

2�2x2 - 5 �2−5�+7x2−5x+7​ = 10$x$ - 17 Đk �2x2 - 5$x$ + 7  ≥ 0

�2x2 - 2.5225​$x$ + 254425​ + 3443​ = ($x$ - 5225​)2 + 3443​ > 0 ∀ $x$

ta có: 2�2x2 - 5�2−5�+7x2−5x+7​ = 10$x$ - 17

2�2x2 - 5�2−5�+7x2−5x+7​ - 10$x$ + 17 = 0

(2�2x2 - 10$x$ + 14)  -  5�2−5�+7x2−5x+7​ + 3 = 0

2.(�2x2 - 5$x$ + 7) - 5.�2−5�+7x2−5x+7​ + 3 = 0

Đặt �2−5�+7x2−5x+7​ = y > 0 ta có: 

2y2 - 5y + 3  = 0

2 + (-5) + 3 = 0

⇒ y1= 1; y2 =  3223​ 

TH1 y = 1 ⇒ �2−5�+7x2−5x+7​  = 1

⇒ �2x2 - 5$x$ + 7  = 1

    �2x2 - 5$x$ + 6 = 0

     $\Delta$ = 25 -  24 = 49

    �1x1​ = −(−5)+122−(−5)+1​​ =  3;

    �2x2​ =  −(−5)−122−(−5)−1​​  = 2;

TH2  y = 3223​

        �2−5�+7x2−5x+7​ = 3223​

         �2x2 - 5$x$ + 7 = 9449​

         4�2x2 - 20$x$ + 28 = 9

          4�2x2 - 20$x$ + 19 = 0

           Δ′Δ′ = 102 - 4.19

          Δ′Δ′ = 24

           �1x1​ = −(−10)+2444−(−10)+24​​ = 10+244410+24​​

           �2x2​ = −(−10)−2444−(−10)−24​​ = 10−244410−24​​

            8 - 566​

Từ các lập luận trên kết luận phương trình có tập nghiệm là:

S = {8 - 566​; 2 ; 3; 8 + 566​}

a, (x+2)(x-3)=0

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

=>S={-2;-3}

b, (x-5)(7-x)=0

\(\left\{{}\begin{matrix}x-5=0\\7-x=0\end{matrix}\right.\left\{{}\begin{matrix}x=5\\-x=-7\end{matrix}\right.\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)

=>S={5;7}

c, (2x+3)(-x+7)=0

\(\left\{{}\begin{matrix}2x+3=0\\-x+7=0\end{matrix}\right.\left\{{}\begin{matrix}2x=-3\\-x=-7\end{matrix}\right.\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=7\end{matrix}\right.\)

=>S={-3/2;7}

a) (x+2)(x+3)=0

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

<=> \(\left\{{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

b) (x-5)(7-x)

<=> \(\left\{{}\begin{matrix}x-5=0\\7-x=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)

c) ( 2x+3)(-2+7)

<=>\(\left\{{}\begin{matrix}2x+3=0\\7-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{2}{7}\end{matrix}\right.\)

d) ( -10x+5)(2x+8)

<=>\(\left\{{}\begin{matrix}5-10x=0\\2x+8=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-4}{1}\end{matrix}\right.\)

e) (x-1)(x+5)(-3x+8)=0

<=> \(\left\{{}\begin{matrix}x-1=0\\x+5=0\\8-3x=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=1\\x=-5\\x=\frac{8}{3}\end{matrix}\right.\)

f) (x-1)(3x+1)=0

<=>\(\left\{{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=\frac{-1}{3}\end{matrix}\right.\)

g) (x-1)(x+2)(x-3)=0

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

<=> \(\left\{{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)

h) (5x+3)(x²+4)(x-1)=0

<=> \(\left\{{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\)

x²+4 > 0 với mọi x∈ R

<=>\(\left\{{}\begin{matrix}x=\frac{-3}{5}\\x=1\end{matrix}\right.\)

Bạn tự kết luận nha , thông cảm cho tớ !