\(6x^4-x^3-7x^2+x+1=0\)

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

\(\Leftrightarrow\left(x-1\right)\left[\left(6x^3-3x^2\right)+\left(8x^2-4x\right)+\left(2x-1\right)\right]=0\)

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

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

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

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

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

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

a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)

\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)

\(\Leftrightarrow3-x-2+7x-7=0\)

\(\Leftrightarrow6x-6=0\)

hay x=1(loại

b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)

\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)

Suy ra: \(-2-x^2+5x-4=x^2+x-6\)

\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow3-x-2+7x-7=0\)

\(\Rightarrow6x-6=0\)

\(\Rightarrow x=1\)

bài này ai kamf chua 

rồi bấy bề

giờ còn mỗi bài cực trị thôi 

đợi mình up thêm nha

a.

a) Chuyển vế và rút gọn được 4x = 16, tìm được x = 4.

b) Đưa PT về dạng 5x = 15, tìm được x = 3

c) Quy đồng, khử mẫu thu được 6x - 9 + 24 = 2 - 2x.

Từ đó tìm được x = - 13 8  

d) Quy đồng, khử mẫu thu được 30x + 9 = 36 + 24x +32

Từ đó tìm được x = 59 6

a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)

=>\(x-6x+18=21-5x-3\)

=>18=18(luôn đúng)

=>\(x\in R\)

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

=>\(x^2-4-x^2+2x-1=2x+2\)

=>2x-5=2x+2

=>-7=0(vô lý)

=>\(x\in\varnothing\)

c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)

=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)

=>3(9x+4)=18-2(3x-5)

=>27x+12=18-6x+10

=>27x+12=-6x+28

=>33x=16

=>\(x=\dfrac{16}{33}\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)

=>6x+1+5x-25=3x-6

=>11x-24=3x-6

=>8x=18

=>\(x=\dfrac{9}{4}\left(nhận\right)\)

a: 

=>

=>18=18(luôn đúng)

=>

b: 

=>

=>2x-5=2x+2

=>-7=0(vô lý)

=>x∈∅

c: 3(9x+4)18=1818−2(3x−5)18

=>3(9x+4)=18-2(3x-5)

=>27x+12=18-6x+10

=>27x+12=-6x+28

=>33x=16

=>

=>

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

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

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

a: =>3x^2-3x-2x+2=0

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

=>x=2/3 hoặc x=1

b: =>2x^2=11

=>x^2=11/2

=>\(x=\pm\dfrac{\sqrt{22}}{2}\)

c: Δ=5^2-4*1*7=25-28=-3<0

=>PTVN

f: =>6x^4-6x^2-x^2+1=0

=>(x^2-1)(6x^2-1)=0

=>x^2=1 hoặc x^2=1/6

=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)

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

=>x=5/2 hoặc x=-5/2

e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2

=>3x^2+5x-8=0 và x>=-1/2

=>3x^2+8x-3x-8=0 và x>=-1/2

=>(3x+8)(x-1)=0 và x>=-1/2

=>x=1