a:

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

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

=>-2x^2+x+1-2x^2+2x=0

=>-4x^2+3x+1=0

=>4x^2-3x-1=0

=>4x^2-4x+x-1=0

=>(x-1)(4x+1)=0

=>x=1(loại) hoặc x=-1/4(nhận)

b: =>2x+6x=x+3(2x+1)

=>x+6x+3=8x

=>7x+3=8x

=>-x=-3

=>x=3(nhận)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\\ \Rightarrow7.\left(2x-1\right)-3.\left(5x+2\right)=21.\left(x+13\right)\\ \Rightarrow14x-7-15x-6=21x+273\\\Rightarrow -x-21x=273+13\\ \Rightarrow-22x=286\\ \Rightarrow x=-13\\ b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}=0\\ \Rightarrow9.\left(x+3\right)+6=4.\left(5x+9\right)-3.\left(7x-9\right)=0\\\Rightarrow 9x+27+6=20x+36-21x+27\\ \Rightarrow9x+33=-x+63\\ \Rightarrow10x=30\\ \Rightarrow x=3\)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Rightarrow7\left(2x-1\right)-3\left(5x+2\right)-21x-273=0\)

\(\Rightarrow14x-7-15x-6-21x-273=0\)

\(\Rightarrow-22x=286\)

\(\Rightarrow x=-13\)

\(b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}\)

\(\Rightarrow9\left(x+3\right)+6-4\left(5x+9\right)+3\left(7x-9\right)=0\)

\(\Rightarrow9x+27+6-20x-36+21x-27=0\)

\(\Rightarrow10x=30\Rightarrow x=3\)

a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)

\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)

=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)

=>30y+25=25y

=>5y=-25

=>y=-5(loại)

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

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=0(nhận) hoặc x=3(loại)

c: =>x^2-9-6(2x+7)=-13(x+3)

=>x^2-9-12x-42+13x+39=0

=>x^2+x-6=0

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

=>x=2(nhận) hoặc x=-3(loại)

a: \(\Leftrightarrow\dfrac{3}{x-2}=\dfrac{2x-1}{x-2}-\dfrac{x\left(x-2\right)}{x-2}\)

=>3=2x-1-x^2+2x

=>3=-x^2+4x-1

=>x^2-4x+1+3=0

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

=>x=2(loại)

b: =>(x+2)(2x-4)=x(2x+3)

=>2x^2-4x+4x-8=2x^2+3x

=>3x=-8

=>x=-8/3(nhận)

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

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

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

=>x=1/2 hoặc x=2

b: =>3(x-1)+4(x+1)=6(x-1)

=>3x-3+4x+4=6x-6

=>7x+1=6x-6

=>x=-7

c: =>x(x-3)-(x+2)(x+3)+16=0

=>x^2-3x-x^2-5x-6+16=0

=>10-8x=0

=>x=5/4

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo để được hỗ trợ tốt hơn. Viết ntn nhìn rất khó đọc

1: =>x-3+3x-9-2(3-x)=60

=>4x-12-6+2x=60

=>6x-18=60

=>6x=78

=>x=13

2: ĐKXĐ: x<>-1; x<>3

heoheo lần sau bạn đánh = kí hiệu đi :(((

a/ \(\dfrac{x}{3}+\dfrac{2x-1}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow2x+2x-1=3\)

<=> 4x = 4 <=> x = 1

Vậy x = 1

b/ \(\dfrac{3x+1}{2}+\dfrac{x-1}{3}=\dfrac{x-9}{6}\)

\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)

\(\Leftrightarrow9x+3+2x-2=x-9\)

\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)

Vậy pt có nghiệm x = -1

c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)

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

\(\Leftrightarrow x^2+2x-x-2=x^2-2x+3x-6\)

\(\Leftrightarrow0x=-4\left(voly\right)\)

Vậy pt vô nghiệm

d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)

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

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

\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)

\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)

Vậy pt có nghiệm x=....

e/ như ý d

a) \(3-2x>4\)

\(\Leftrightarrow-2x>1\)

\(\Leftrightarrow x< \frac{-1}{2}\)

b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)

\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow-4x-13-18x+54=x^2-9\)

\(\Leftrightarrow x^2+22x-50=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)

\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )

Vậy....

\(a,\)\(3-2x>4\)

\(\Rightarrow-2x>1\)

\(\Rightarrow x< \frac{-1}{2}\)