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

=>x^2+2x=x

=>x^2+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: =>-9(5x-8)+4(7x-12)=-6(x+18)

=>-45x+72+28x-48=-6x-108

=>-17x+24=-6x-108

=>-11x=-132

=>x=12

a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)

\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)

hay \(x=-\dfrac{41}{12}\)

Vậy: \(x=-\dfrac{41}{12}\)

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)

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

`<=> 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/ \(2x+\frac{1}{7}=\frac{1}{3}\)

=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)

=> \(2x=\frac{4}{21}\)

=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)

b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)

=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)

=> \(x-\frac{1}{2}=\frac{4}{27}\)

=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)

c/ \(\left(x-5\right)^2+4=68\)

=> \(\left(x-5\right)^2=68-4=64\)

=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)

d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)

=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)

e) \(5x+2=3x+8\)

=> \(5x-3x=8-2=6\)

=> \(2x=6\)

=> \(x=6:2=3\)

f/ \(26-\left(5-2x\right)=27\)

=> \(5-2x=26-27=-1\)

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

=> \(x=6:2=3\)

g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)

=> \(4x-8-2x+6=4\)

=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)

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

=> \(2x=4+2=6\)

=> \(x=6:2=3\)

h/ \(\left(x+3\right)^3:3-1=-10\)

=> \(\left(x+3\right)^3:3=-10+1=-9\)

=> \(\left(x+3\right)^3=-9.3=-27\)

=> \(x+3=-3\)

=> \(x=-3-3=-6\)

`a,5x-2=3x+1`

`<=>5x-3x=1+2`

`<=>2x=3`

`<=>x=3/2`

Vậy `x=3/2`

`b,(x+5)(2x-3)=0`

`<=>` $\left[ \begin{array}{l}x+5=0\\2x-3=0\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-5\end{array} \right.$

Vậy `S={-5,3/2}`

Dẫn khí CO dư qua ống sứ chứa 20(g) FexOy và PbO cho đến khi phản ứng kết thúc, thu được chất rắn A và khí B. Cho A vào dung dịch HCl dư, thấy thoát ra 4,48(l) khí ở đktc. Dẫn khí B vào dung dịch Ca(OH)2 dư thu được 35(g) kết tủa. Lập CT FexOy                

giup minh giai voi

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)

giải nhanh giup mình nhé