a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

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

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

PT \(\Leftrightarrow9x^2-6x+1-9x+6=9x^2-18x-27\)

\(\Leftrightarrow9x^2-6x+1-9x+6-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy ...

Ta có: \(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-3x+x-3\right)\)

\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)

\(\Leftrightarrow9x^2-15x+7-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow3x=-34\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy: \(S=\left\{-\dfrac{34}{3}\right\}\)

a: =>x(x+3)=0

=>x=0 hoặc x=-3

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

=>x=0 hoặc x=1/2

c: =>(x-7)(2x+3-x)=0

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

=>x=7 hoặc x=-3

d: =>(x-2)(3x-1-x-3)=0

=>(x-2)(2x-4)=0

=>x=2

a)

`x^2 +3x=0`

`<=>x(x+3)=0`

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

b)

`x-2x^2 =0`

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

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

c)

`(x-7)(2x+3)=x(x-7)`

`<=>(x-7)(2x+3)-x(x-7)=0`

`<=>(x-7)(2x+3-x)=0`

`<=>(x-7)(x+3)=0`

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

d)

`(x-2)(x+3)=(x-2)(3x-1)`

`<=>(x-2)(x+3)-(x-2)(3x-1)=0`

`<=>(x-2)(x+3-3x+1)=0`

`<=>(x-2)(-2x+4)=0`

\(< =>\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\\ < =>x=2\)

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

<=>x(x+3)=0

x=0 hoặc x+3=0

x=0 hoặc x=-3

b)x-2x²=0

x(1-2x)=0

x=0 hoặc 1-2x=0

x=0 hoặc x=0,5

c)(x-7)(2x+3)=4(x-7)

(x-7)(2x+3)-4(x-7)=0

(x-7)(2x+3-4)=0

x-7=0 hoặc 2x+3-4=0

x=7 hoặc x=0,5

d)(x-2)(x+3)=(x-2)(3x-1)

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

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

x-2=0 hoặc x+3-3x+1=0

x=2 hoặc x-3x=-3-1

               -2x=-4

               x=2

ĐKXĐ: \(x\notin\left\{2;-1;\dfrac{-3\pm\sqrt{17}}{2}\right\}\)

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

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

=>\(\dfrac{x^3+3x^2-2x+3x^3-3x^2-6x}{\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2}=1\)

=>\(4x^3-8x=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)

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

=>\(\left(x^2-2\right)^2-2x\left(x^2-2\right)-3x^2=0\)

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

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

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

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

=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=1\left(nhận\right)\\x=\dfrac{3\pm\sqrt{17}}{2}\left(nhận\right)\end{matrix}\right.\)

1) Ta có: \(4x+8=3x-1\)

\(\Leftrightarrow4x-3x=-1-8\)

\(\Leftrightarrow x=-9\)

2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)

\(\Leftrightarrow10-5x-15-3x+3>0\)

\(\Leftrightarrow-8x>2\)

hay \(x< \dfrac{-1}{4}\)

=>(9x^2+24x-6x-16)(x^2+2x+1)=-16

=>(9x^2+18x-16)(x^2+2x+1)=-16

=>(9x^2+18x+9-25)(x^2+2x+1)=-16

=>[9(x+1)^2-25](x+1)^2=-16

=>9(x+1)^4-25(x+1)^2+16=0

Đặt (x+1)^2=a

=>9a^2-25a+16=0

=>a=1 hoặc a=16/9

=>(x+1)^2=1 hoặc (x+1)^2=16/9

=>\(x\in\left\{0;-2;\dfrac{1}{3};-\dfrac{7}{3}\right\}\)

CẢM ƠN NHÌU NHA