a: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow6x^2-3x-4x+2-6x^2+3x=0\)

\(\Leftrightarrow-4x=-2\)

hay \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x=x^3-x^3-1\)

hay x=-1

1) 2x.(5x-3x)+2x.(3x-5)-3.(x-7)=3

   10x-6x^2+6x^2-10x-3x+21=3

    -3x                             =-18

suy ra x=6

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

     3x^2 +3x-2x^2-4x =-1-x

     x^2 =-1

suy ra không có giá trị nào của x thỏa mãn đề bài

3) 2x^2 +3.(x^2-1)=5x(x+1)

  2x^2 +3x^2-3 =5x^2+5x

  -5x      =3

x=-3/5

giải rồi đấy

nhớ tích đúng nha :)

bạn coi lại đề câu 1 đi

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

`@` `\text {Ans}`

`\downarrow`

Mình làm lại nhé ;-;

`@`` \text {Ans}`

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Sửa lại đúng lớp đi c nhé.

đây ak  

chiu

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

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

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

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

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

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

Vậy \(S=\left\{0;5\right\}\)

\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)

\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)

\(5x-x=3x+7\)

\(4x-3x=7\)

\(x=7\)

(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7

=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7

=>-x^2-x+x^2+5x=3x+7

=>4x=3x+7

=>x=7