1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)

1: =>|x-4|+|x+2|=0

=>x-4=0 và x+2=0

=>\(x\in\varnothing\)

2: =>x^2-x-6=3x+5

=>x^2-4x-11=0

=>x^2-4x+4-15=0

=>(x-2)^2-15=0

=>x=căn 15+2 hoặc x=-căn 15+2

3: =>x^2-x=3x+5

=>x^2-4x-5=0

=>(x-5)(x+1)=0

=>x=-1 hoặc x=5

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a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

\(\left(x+4\right)\left(3x-1\right)+\left(x^2+8x+16\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+4\right)\left(3x-1\right)=0\\x^2+8x+16=0\end{cases}}\)

Xét PT 1 : \(\left(x+4\right)\left(3x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{1}{3}\end{cases}}\)

Xét PT 2 : \(x^2+8x+16=0\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

\(x=\frac{-3}{4}\)hoặc \(x=-4\)

các bạn giúp tóe nhoe

= (3x + 1)^2 - (x-4)^2

= (3x+ 1 +x-4) (3x+1-x+4)

= (4x-3) (2x+5)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

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

3x( x² - 8x + 16 ) = 0

\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2-8x+16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-4\right)^2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=4\end{cases}}.}\)

       3x( x² - 8x + 16 ) = 0

<=> 3x( x² - 2.x.4 + 4² ) = 0

<=> 3x( x - 4 )² = 0

<=> 3x = 0 hoặc ( x - 4 )² = 0

<=> x = 0 hoặc x - 4 = 0

<=> x = 0 hoặc x = 4

KL