a: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>\(\left(x+3\right)\cdot\left(27x^2-12x\right)=0\)

=>3x(x+3)(9x-4)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

b: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

=>\(\left(x-2\right)\left(3x+5\right)=2\left(x-2\right)\left(x+1\right)\)

=>(x-2)(3x+5)-(x-2)(2x+2)=0

=>(x-2)(3x+5-2x-2)=0

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

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

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

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

=>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)

a, \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

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

Vậy tập nghiệm phương trình là S = { 1 ; 2 ; 3 ; 4 } 

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

Vậy tập nghiệm của phương trình là S = { -2 ; 3 } 

c, \(\left|2x-1\right|=\left|x+5\right|\Leftrightarrow\left(2x-1\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(2x-1\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-1-x-5\right)\left(2x-1+x+5\right)=0\Leftrightarrow x=6;x=-\dfrac{4}{3}\)

Vậy tập nghiệm của phương trình là S = { -4/3 ; 6 } 

d, \(\left|3x+1\right|=x-2\)

TH1 : \(3x+1=x-2\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)

TH2 : \(3x+1=-x+2\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

Vậy tập nghiệm của phương trình là S = { -3/2 ; 1/4 } 

các ý còn lại tương tự 

a) Ta có: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)

Vậy: S={3;4;1;2}

a: \(x\left(x-1\right)+2x^2-2=0\)

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

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

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

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

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

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

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

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

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

a: 

=>

=>

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

=>

b: 

=>

=>

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

=>

Đề bài thiếu bạn ạ

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

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

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

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

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

a: \(2x^3-50x=0\)

=>\(2x\left(x^2-25\right)=0\)

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

=>x∈{0;5;-5}

b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

=>2x(3x-5)+(3x-5)=0

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

=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)

c: \(9\left(3x-2\right)=x\left(2-3x\right)\)

=>9(3x-2)-x(2-3x)=0

=>9(3x-2)+x(3x-2)=0

=>(3x-2)(x+9)=0

=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)

d: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

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

=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

e: \(25x^2-2=0\)

=>\(25x^2=2\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

f: \(x^2-25=6x-9\)

=>\(x^2-6x-16=0\)

=>(x-8)(x+2)=0

=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)

g: 5x(x-3)-2x+6=0

=>5x(x-3)-2(x-3)=0

=>(x-3)(5x-2)=0

=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)

h: 3x(x-7)-2(x-7)=0

=>(x-7)(3x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)

i: \(7x^2-28=0\)

=>\(7x^2=28\)

=>\(x^2=4\)

=>x=2 hoặc x=-2

j: 2x+1+x(2x+1)=0

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

=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)

k: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

=>4(x+2)=0

=>x+2=0

=>x=-2

l: \(x^3+5x^2-4x-20=0\)

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

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

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

=>x∈{-5;2;-2}

m: \(x^2-25+2\left(x+5\right)=0\)

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

=>(x+5)(x-3)=0

=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)

n: \(x^2-3x+2=0\)

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

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

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

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

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

=>\(\left(x-2\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

p: \(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

=>(x-7)(x+2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)

q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

=>\(x^2-4x+4-x^2+9=6\)

=>-4x+13=6

=>-4x=6-13=-7

=>x=7/4

r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)

=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)

=>-4x+26=18