19 22 25 28 5(3x + 2) – 4(2x +3) x*(1 + 2x) 4(1 + x) – 3(2x-5) 4x–8(6) - X) 23/ ... 2x” – 4x + 3x – 6 = 2x” – X-6 (b) (x-3) = (x-3)(x-3) (c) (2x+y)(2x–y) = x* = x* – 3x ... (x - 6)” 7 (3x + 5)(x-6) 8 (8x + 2)(3x + 4) (4x – 1)(2x – 3) 10 (2x +5)* 11 (8x – 3)(2x + ... 27 (4x + 3y)(x + y) 28 (2x + 5)(5x – 2) (4x – 3y)(4x + y) 30 (7x + 2y)(3x + 4y) 24/ ...

\(a)=3x\cdot\left(2x-7-4x+5\right)=3x\cdot\left(-2x-2\right)=3x\cdot\left[-2\cdot\left(x+1\right)\right]\)

a, \(15^4-12x^3+9x^2\)

b,\(-15x^3y^2+25x^2y^2-5xy^3\)

c, \(5x^3-19x^2+12x\)

d, 

x3+xy2+5x2y−9x2y−3y3−15xy2=3x3−3y3−14xy2−4x2y

\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3+xy^2+5x^2y-9x^2y-3y^3-15xy^2=3x^3-14xy^2-4x^2y-3y^3\)

\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-19x^2+12x\\ d,=3x^3+xy^2+5x^2y-9x^2y-3y^3-15xy^2\\ =3x^3-3y^3-14xy^2-4x^2y\)

a) 5x^2. (3x^2 - 7x + 2)

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

= 15x^4 - 35x^3 + 10x^2

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

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

= 10x^4 - 25x^3 + 17x^2 - 3x

a) \(\left\{{}\begin{matrix}2x+3y=-5\\6x-5y=27\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+9y=-15\\6x-5y=27\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}14y=-42\\2x+3y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x+3.\left(-3\right)=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-9=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

b) \(3x^2+4x=0\) 

\(\Leftrightarrow x\left(3x+4\right)=0\)

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

Vậy phương trình có tập nghiệm là: \(S=\left\{0;-\dfrac{4}{3}\right\}\)

c) Đặt:  \(x^2=t\left(t\ge0\right)\)

\(\Rightarrow\) Ta có phương trình mới:

\(t^2-3t-4=0\) 

Ta có: a - b + c = 1 + 3 - 4 = 0

\(\Rightarrow t_1=-1\left(loại\right);t_2=4\left(TM\right)\)

\(\Rightarrow x=\pm2\)

Vậy tập nghiệm của phương trình là: S = {2; -2}

a, \(\left\{{}\begin{matrix}2x+3y=-5\\6x-5y=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+9y=-15\left(1\right)\\6x-5y=27\left(2\right)\end{matrix}\right.\)

Lấy (1) - (2) ta được : \(14y=-15-27=-42\Leftrightarrow y=-3\)

\(\Rightarrow6x-27=-15\Leftrightarrow6x=12\Leftrightarrow x=2\)

Vậy \(\left(x;y\right)=\left(2;-3\right)\)

b, \(3x^2+4x=0\Leftrightarrow x\left(3x+4\right)=0\Leftrightarrow x=0;x=-\dfrac{4}{3}\)

c, \(x^4-3x^2-4=0\Leftrightarrow x^4+x^2-4x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)+x^2-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow x=\pm2;x^2+1>0\)

Vậy nghiệm của phương trình là x = -2 ; x = 2 

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

\(1,=3x^2-6x+x-2=3x^2-5x-2\\ 2,??\\ 3,=3x^3y^2:3xy+6x^2y^3:3xy-12xy^4:3xy=x^2y+2xy^2-4y^3\\ 4,=3x^3y^2:4xy+6x^2y^3:4xy-12xy^4:4xy\\ =\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-3x^3\\ 5,\left(2x^3-5x^2+7x-6\right):\left(2x-3\right)=x^2-x+2\\ 6,\left(x^4-x^3+3x^2+x+2\right):\left(x^2-1\right)=x^2-x+4\left(dư6\right)\) 

1: =3x^2+x-6x-2=3x^2-5x-2

3: =x^2y+2xy^2-4y^3

4: =3/4x^2y+3/2xy^2-3y^3

5: \(=\dfrac{2x^3-3x^2-2x^2+3x+4x-6}{2x-3}=x^2-x+2\)