a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

tham khảo

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

a,Để \(|2x+1|+|x-2|=0\Leftrightarrow\hept{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)(vô lý)

=> ko có x thỏa mãn

b,\(|x+5|=2x-1\Leftrightarrow1-2x< x+5< 2x-1\)

giai gium mik cau c 

...... mk ko bít làm

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

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

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

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

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

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

b: Ta có: \(5\left(x-1\right)^2-\left(1-x\right)\)

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

\(=\left(x-1\right)\left(5x-5+1\right)\)

\(=\left(x-1\right)\left(5x-4\right)\)

a: Ta có: \(5x^2-4xy-x^2y\)

\(=x\left(5x-4y-xy\right)\)

\(a,\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\\ \Leftrightarrow4x^2+x-12x-3-\left(4x^2-28x-x+7\right)-15=0\\ \Leftrightarrow4x^2-11x-3-4x^2+29x-7-15=0\\ \Leftrightarrow18x=25\\ \Leftrightarrow x=\dfrac{25}{18}\)

Vậy \(x=\dfrac{25}{18}\)

\(b,\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-3\right)=4\\ \Leftrightarrow x^3+1-x^3+3x-4=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow x=1\)

Vậy \(x=1\)

\(c,\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)-6x=0\\ \Leftrightarrow x^3-27+5x-x^3-6x=0\\ \Leftrightarrow-x-27=0\\ \Leftrightarrow x=-27\)

Vậy \(x=-27\)

\(d,\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\\ \Leftrightarrow25x^2-1-25x^2+7x-15=0\\ \Leftrightarrow7x-16=0\\ \Leftrightarrow x=\dfrac{16}{7}\)

Vậy \(x=\dfrac{16}{7}\)

=2x^3-2x^2-5x-10-2x^2+4x+x^2(2x-3)-x(x+1)-3x+2

=2x^3-4x^2-4x-8+2x^3-6x^2-x^2+x

=4x^3-11x^2-3x-8