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)

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

a: =>3(3x-7)+2(x+1)=-96

=>9x-21+2x+2=-96

=>11x-19=-96

=>11x=-96+19=-75

=>x=-75/11

b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)

=>15x-5(x+1)=3(2x+1)

=>15x-5x-5=6x+3

=>10x-5=6x+3

=>4x=8

=>x=2

a)

\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)

\(< =>9x-21+2x+2=-96\)

\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)

b)

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

\(a,\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\\ \Rightarrow3\left(3x-7\right)+2\left(x+1\right)=2.3.\left(-16\right)\\ \Rightarrow9x-21+2x+2=-96\\ \Rightarrow11x-19=-96\\ \Rightarrow11x=-77\\ \Rightarrow x=-7\\ b,x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\\ \Rightarrow15x-5.\left(x+1\right)=3.\left(2x+1\right)\\\Rightarrow 15x-5x-5=6x+3\\ \Rightarrow10x-6x=3+5\\ \Rightarrow4x=8\\ \Rightarrow x=2\)

a)

\(\dfrac{x-2}{4}+\dfrac{2x-3}{3}=\dfrac{x-18}{6}\)

`<=> 3x-6+8x-12=2x-36`

`<=> 3x+8x-2x=-36+6+12`

`<=> 9x=-18`

`<=> x=-2`

b)

\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\left(x\ne3;x\ne-3\right)\)

suy ra

`(x+3)^2 +(3-x)(x-3)=36`

`<=>x^2 +6x+9+3x-9-x^2 +3x=36`

`<=> x^2 -x^2 +6x+3x+3x+9-9-36=0`

`<=> 12x-36=0`

`<=> 12x=36`

`<=> x=3 (KTMĐK)

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

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

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

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

\(\Rightarrow-22x=286\)

\(\Rightarrow x=-13\)

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

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

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

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

\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)

\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)

\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)

\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

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: =>(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